This is a very simple proof. To make the notation a little clearer let’s define the function \(f\left( x \right) = c\) then what we’re being asked to prove is that \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = c\). So let’s do that.

Let \(\varepsilon > 0\) and we need to show that we can find a \(\delta > 0\) so that

\[\left| {f\left( x \right) - c} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

The left inequality is trivially satisfied for any \(x\) however because we defined \(f\left( x \right) = c\). So simply choose \(\delta > 0\) to be any number you want (you generally can’t do this with these proofs). Then,

\[\left| {f\left( x \right) - c} \right| = \left| {c - c} \right| = 0 < \varepsilon \]

There are several ways to prove this part. If you accept 3 And 7 then all you need to do is let \(g\left( x \right) = c\) and then this is a direct result of 3 and 7. However, we’d like to do a more rigorous mathematical proof. So here is that proof.

First, note that if \(c = 0\) then \(cf\left( x \right) = 0\) and so,

\[\mathop {\lim }\limits_{x \to a} \left[ {0f\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} 0 = 0 = 0f\left( x \right)\]

The limit evaluation is a special case of 7 (with \(c = 0\)) which we just proved Therefore we know 1 is true for \(c = 0\) and so we can assume that \(c \ne 0\) for the remainder of this proof.

Let \(\varepsilon > 0\) then because \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) by the definition of the limit there is a \({\delta _{\,1}} > 0\) such that,

\[\left| {f\left( x \right) - K} \right| < \frac{\varepsilon }{{\left| c \right|}}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,1}}\]

Now choose \(\delta = {\delta _{\,1}}\) and we need to show that

\[\left| {cf\left( x \right) - cK} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

and we’ll be done. So, assume that \(0 < \left| {x - a} \right| < \delta \)and then,

\[\left| {cf\left( x \right) - cK} \right| = \left| c \right|\left| {f\left( x \right) - K} \right| < \left| c \right|\frac{\varepsilon }{{\left| c \right|}} = \varepsilon \]

We’ll be doing this proof in two parts. First let’s prove \(\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = K + L\).

Let \(\varepsilon > 0\) then because \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) there is a \({\delta _{\,1}} > 0\) and a \({\delta _2} > 0\) such that,

\[\begin{align*}\left| {f\left( x \right) - K} \right| & < \frac{\varepsilon }{2}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta {\,_1}\\ \left| {g\left( x \right) - L} \right|& < \frac{\varepsilon }{2}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}\end{align*}\]

Now choose \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\). Then we need to show that

\[\left| {f\left( x \right) + g\left( x \right) - \left( {K + L} \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

Assume that \(0 < \left| {x - a} \right| < \delta \). We then have,

\[\begin{align*}\left| {f\left( x \right) + g\left( x \right) - \left( {K + L} \right)} \right| & = \left| {\left( {f\left( x \right) - K} \right) + \left( {g\left( x \right) - L} \right)} \right|\\ & \le \left| {f\left( x \right) - K} \right| + \left| {g\left( x \right) - L} \right|\hspace{0.25in}{\mbox{by the triangle inequality}}\\ & < \frac{\varepsilon }{2} + \frac{\varepsilon }{2}\\ & = \varepsilon \end{align*}\]

In the third step we used the fact that, by our choice of \(\delta \), we also have \(0 < \left| {x - a} \right| < {\delta _{\,1}}\) and \(0 < \left| {x - a} \right| < {\delta _{\,2}}\) and so we can use the initial statements in our proof.

Next, we need to prove \(\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] = K - L\). We could do a similar proof as we did above for the sum of two functions. However, we might as well take advantage of the fact that we’ve proven this for a sum and that we’ve also proven 1.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + \left( { - 1} \right)g\left( x \right)} \right]\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} \left( { - 1} \right)g\left( x \right)\hspace{0.25in}{\mbox{by first part of }}{\bf{2}}{\mbox{.}}\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \left( { - 1} \right)\mathop {\lim }\limits_{x \to a} g\left( x \right)\hspace{0.25in}{\mbox{by }}{\bf{1}}{\mbox{.}}\\ & = K + \left( { - 1} \right)L\\ & = K - L\end{align*}\]

This one is a little tricky. First, let’s note that because \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = K\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) we can use 2 and 7 to prove the following two limits.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right] & = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} K = K - K = 0\\ \mathop {\lim }\limits_{x \to a} \left[ {g\left( x \right) - L} \right] & = \mathop {\lim }\limits_{x \to a} g\left( x \right) - \mathop {\lim }\limits_{x \to a} L = L - L = 0\end{align*}\]

Now, let \(\varepsilon > 0\). Then there is a \({\delta _{\,1}} > 0\) and a \({\delta _2} > 0\) such that,

\[\begin{align*}\left| {\left( {f\left( x \right) - K} \right) - 0} \right| & < \sqrt \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,1}}\\ \left| {\left( {g\left( x \right) - L} \right) - 0} \right| & < \sqrt \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}\end{align*}\]

Choose \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\). If \(0 < \left| {x - a} \right| < \delta \) we then get,

\[\begin{align*}\left| {\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] - 0} \right| & = \left| {f\left( x \right) - K} \right|\left| {g\left( x \right) - L} \right|\\ & < \sqrt \varepsilon \sqrt \varepsilon \\ & = \varepsilon \end{align*}\]

So, we’ve managed to prove that,

\[\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] = 0\]

This apparently has nothing to do with what we actually want to prove, but as you’ll see in a bit it is needed.

Before launching into the actual proof of 3 let’s do a little Algebra. First, expand the following product.

\[\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] = f\left( x \right)g\left( x \right) - Lf\left( x \right) - Kg\left( x \right) + KL\]

Rearranging this gives the following way to write the product of the two functions.

\[f\left( x \right)g\left( x \right) = \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + Lf\left( x \right) + Kg\left( x \right) - KL\]

With this we can now proceed with the proof of 3.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} f\left( x \right)g\left( x \right) & = \mathop {\lim }\limits_{x \to a} \left[ {\left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + Lf\left( x \right) + Kg\left( x \right) - KL} \right]\\ & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - K} \right]\left[ {g\left( x \right) - L} \right] + \mathop {\lim }\limits_{x \to a} Lf\left( x \right) + \mathop {\lim }\limits_{x \to a} Kg\left( x \right) - \mathop {\lim }\limits_{x \to a} KL\\ & = 0 + \mathop {\lim }\limits_{x \to a} Lf\left( x \right) + \mathop {\lim }\limits_{x \to a} Kg\left( x \right) - \mathop {\lim }\limits_{x \to a} KL\hspace{0.5in}\hspace{0.5in}\\ & = LK + KL - KL\\ & = LK\end{align*}\]

Fairly simple proof really, once you see all the steps that you have to take before you even start. The second step made multiple uses of property 2. In the third step we used the limit we initially proved. In the fourth step we used properties 1 and 7. Finally, we just did some simplification.

This one is also a little tricky. First, we’ll start of by proving,

\[\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}} = \frac{1}{L}\]

Let \(\varepsilon > 0\) . We’ll not need this right away, but these proofs always start off with this statement. Now, because \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\) there is a \({\delta _{\,1}} > 0\) such that,

\[\left| {g\left( x \right) - L} \right| < \frac{{\left| L \right|}}{2}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,1}}\]

Now, assuming that \(0 < \left| {x - a} \right| < {\delta _{\,1}}\) we have,

\[\begin{align*}\left| L \right| & = \left| {L - g\left( x \right) + g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{just adding zero to }}L\\ & \le \left| {L - g\left( x \right)} \right| + \left| {g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{using the triangle inequality}}\\ & = \left| {g\left( x \right) - L} \right| + \left| {g\left( x \right)} \right| & \hspace{0.5in} & \left| {L - g\left( x \right)} \right| = \left| {g\left( x \right) - L} \right|\\ & < \frac{{\left| L \right|}}{2} + \left| {g\left( x \right)} \right| & \hspace{0.5in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < {\delta _1}\end{align*}\]

Rearranging this gives,

\[\left| L \right| < \frac{{\left| L \right|}}{2} + \left| {g\left( x \right)} \right|\hspace{0.25in} \Rightarrow \hspace{0.25in}\frac{{\left| L \right|}}{2} < \left| {g\left( x \right)} \right|\hspace{0.5in} \Rightarrow \hspace{0.25in}\frac{1}{{\left| {g\left( x \right)} \right|}} < \frac{2}{{\left| L \right|}}\]

Now, there is also a \({\delta _2} > 0\) such that,

\[\left| {g\left( x \right) - L} \right| < \frac{{{{\left| L \right|}^2}}}{2}\varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < {\delta _{\,2}}\]

Choose \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\). If \(0 < \left| {x - a} \right| < \delta \) we have,

\[\begin{align*}\left| {\frac{1}{{g\left( x \right)}} - \frac{1}{L}} \right| & = \left| {\frac{{L - g\left( x \right)}}{{Lg\left( x \right)}}} \right| & \hspace{0.5in} & {\mbox{common denominators}}\\ & = \frac{1}{{\left| {Lg\left( x \right)} \right|}}\left| {L - g\left( x \right)} \right| & \hspace{0.25in} & {\mbox{doing a little rewriting}}\\ & = \frac{1}{{\left| L \right|}}\frac{1}{{\left| {g\left( x \right)} \right|}}\left| {g\left( x \right) - L} \right| & \hspace{0.25in} & {\mbox{doing a little more rewriting}}\\ & < \frac{1}{{\left| L \right|}}\frac{2}{{\left| L \right|}}\left| {g\left( x \right) - L} \right| & \hspace{0.25in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < \delta \le {\delta _1}\\ & < \frac{2}{{{{\left| L \right|}^2}}}\frac{{{{\left| L \right|}^2}}}{2}\varepsilon & \hspace{0.5in} & {\mbox{assuming that }}0 < \left| {x - a} \right| < \delta \le {\delta _2}\\ & = \varepsilon \end{align*}\]

Now that we’ve proven \(\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}} = \frac{1}{L}\) the more general fact is easy.

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right)\frac{1}{{g\left( x \right)}}} \right]\\ & = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} \frac{1}{{g\left( x \right)}}\hspace{0.5in}{\mbox{using property }}{\bf{3}}{\mbox{.}}\\ & = K\frac{1}{L} = \frac{K}{L}\end{align*}\]

As noted we’re only going to prove 5 for integer exponents. This will also involve proof by induction so if you aren’t familiar with induction proofs you can skip this proof.

So, we’re going to prove,

\[\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} = {\left[ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right]^n} = {K^n},\hspace{0.5in}n \ge 2,\,\,n{\mbox{ is an integer}}{\mbox{.}}\]

For \(n = 2\) we have nothing more than a special case of property 3.

\[\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^2} = \mathop {\lim }\limits_{x \to a} f\left( x \right)f\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right)\mathop {\lim }\limits_{x \to a} f\left( x \right) = KK = {K^2}\]

So, 5 is proven for \(n = 2\). Now assume that 5 is true for \(n - 1\), or \(\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{n - 1}} = {K^{n - 1}}\). Then, again using property 3 we have,

\[\begin{align*}\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} & = \mathop {\lim }\limits_{x \to a} \left( {{{\left[ {f\left( x \right)} \right]}^{n - 1}}f\left( x \right)} \right)\\ & = \mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{n - 1}}\mathop {\lim }\limits_{x \to a} f\left( x \right)\\ & = {K^{n - 1}}K\\ & = {K^n}\end{align*}\]

As pointed out in the Limit Properties section this is nothing more than a special case of the full version of 5 and the proof is given there and so is the proof is not give here.

This is a simple proof. If we define \(f\left( x \right) = x\) to make the notation a little easier, we’re being asked to prove that \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = a\).

Let \(\varepsilon > 0\) and let \(\delta = \varepsilon \). Then, if \(0 < \left| {x - a} \right| < \delta = \varepsilon \) we have,

\[\left| {f\left( x \right) - a} \right| = \left| {x - a} \right| < \delta = \varepsilon \]

So, we’ve proved that \(\mathop {\lim }\limits_{x \to a} x = a\).

This is just a special case of property 5 with \(f\left( x \right) = x\) and so we won’t prove it here.

We will prove \(\mathop {\lim }\limits_{x \to c} \left[ {f\left( x \right) + g\left( x \right)} \right] = \infty \) here. The proof of \(\mathop {\lim }\limits_{x \to c} \left[ {f\left( x \right) - g\left( x \right)} \right] = \infty \) is nearly identical and is left to you.

Let \(M > 0\) then because we know \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) there exists a \({\delta _{\,1}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) we have,

\[f\left( x \right) > M - L + 1\]

Also, because we know \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) there exists a \({\delta _{\,2}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have,

\[0 < \left| {g\left( x \right) - L} \right| < 1\hspace{0.25in}\,\, \to \hspace{0.25in} - 1 < g\left( x \right) - L < 1\hspace{0.25in} \to \,\,\,\,\,\,\,L - 1 < g\left( x \right) < L + 1\]

Now, let \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) and so if \(0 < \left| {x - c} \right| < \delta \) we know from the above statements that we will have both,

\[f\left( x \right) > M - L + 1\hspace{0.5in}\hspace{0.25in}g\left( x \right) > L - 1\]

This gives us,

\[\begin{align*}f\left( x \right) + g\left( x \right) & > M - L + 1 + L - 1\\ & = M\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.5in}f\left( x \right) + g\left( x \right) > M\end{align*}\]

So, we’ve proved that \(\mathop {\lim }\limits_{x \to c} \left[ {f\left( x \right) + g\left( x \right)} \right] = \infty \).

Let \(M > 0\) then because we know \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) there exists a \({\delta _{\,1}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) we have,

\[f\left( x \right) > \frac{{2M}}{L}\]

Also, because we know \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) there exists a \({\delta _{\,2}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have,

\[0 < \left| {g\left( x \right) - L} \right| < \frac{L}{2}\hspace{0.25in} \to \hspace{0.25in} - \frac{L}{2} < g\left( x \right) - L < \frac{L}{2}\hspace{0.25in} \to \hspace{0.25in} \frac{L}{2} < g\left( x \right) < \frac{{3L}}{2}\]

Note that because we know that \(L > 0\) choosing \(\frac{L}{2}\) in the first inequality above is a valid choice because it will also be positive as required by the definition of the limit.

Now, let \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) and so if \(0 < \left| {x - c} \right| < \delta \) we know from the above statements that we will have both,

\[f\left( x \right) > \frac{{2M}}{L}\hspace{0.5in}\hspace{0.25in}g\left( x \right) > \frac{L}{2}\]

This gives us,

\[\begin{align*}f\left( x \right)g\left( x \right) & > \left( {\frac{{2M}}{L}} \right)\left( {\frac{L}{2}} \right)\\ & = M\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.5in}f\left( x \right)g\left( x \right) > M\end{align*}\]

So, we’ve proved that \(\mathop {\lim }\limits_{x \to c} f\left( x \right)g\left( x \right) = \infty \).

Let \(M > 0\) then because we know \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) there exists a \({\delta _{\,1}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) we have,

\[f\left( x \right) > \frac{{ - 2M}}{L}\]

Note that because \(L < 0\) in the case we will have \(\frac{{ - 2M}}{L} > 0\) here.

Next, because we know \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) there exists a \({\delta _{\,2}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have,

\[0 < \left| {g\left( x \right) - L} \right| < - \frac{L}{2}\hspace{0.25in}\,\, \to \hspace{0.25in}\frac{L}{2} < g\left( x \right) - L < - \frac{L}{2}\hspace{0.25in} \to \hspace{0.25in} \frac{{3L}}{2} < g\left( x \right) < \frac{L}{2}\]

Again, because we know that \(L < 0\) we will have \( - \frac{L}{2} > 0\). Also, for reasons that will shortly be apparent, multiply the final inequality by a minus sign to get,

\[\, - \frac{L}{2} < - g\left( x \right) < - \frac{{3L}}{2}\]

Now, let \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) and so if \(0 < \left| {x - c} \right| < \delta \) we know from the above statements that we will have both,

\[f\left( x \right) > \frac{{ - 2M}}{L}\hspace{0.5in}\hspace{0.25in} - g\left( x \right) > - \frac{L}{2}\]

This gives us,

\[\begin{align*} - f\left( x \right)g\left( x \right) & = f\left( x \right)\left[ { - g\left( x \right)} \right]\\ & > \left( { - \frac{{2M}}{L}} \right)\left( { - \frac{L}{2}} \right)\\ & = M\hspace{0.5in}\hspace{0.25in} \Rightarrow \hspace{0.5in} - f\left( x \right)g\left( x \right) > M\end{align*}\]

This may seem to not be what we needed however multiplying this by a minus sign gives,

\[f\left( x \right)g\left( x \right) < - M\]

and because we originally chose \(M > 0\) we have now proven that \(\mathop {\lim }\limits_{x \to c} f\left( x \right)g\left( x \right) = - \infty \).

We’ll need to do this in three cases. Let’s start with the easiest case.

Case 1 : \(L = 0\)

Let \(\varepsilon > 0\) then because we know \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) there exists a \({\delta _{\,1}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) we have,

\[f\left( x \right) > \frac{1}{{\sqrt \varepsilon }} > 0\]

Next, because we know \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = 0\) there exists a \({\delta _{\,2}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have,

\[0 < \left| {g\left( x \right)} \right| < \sqrt \varepsilon \]

Now, let \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) and so if \(0 < \left| {x - c} \right| < \delta \) we know from the above statements that we will have both,

\[f\left( x \right) > \frac{1}{{\sqrt \varepsilon }}\hspace{0.5in}\hspace{0.25in}\left| {g\left( x \right)} \right| < \sqrt \varepsilon \]

This gives us,

\[\left| {\frac{{g\left( x \right)}}{{f\left( x \right)}}} \right| = \frac{{\left| {g\left( x \right)} \right|}}{{f\left( x \right)}} < \frac{{\sqrt \varepsilon }}{{f\left( x \right)}} < \frac{{\sqrt \varepsilon }}{{{}^{1}/{}_{{\sqrt \varepsilon }}}} = \varepsilon \]

In the second step we could remove the absolute value bars from \(f\left( x \right)\) because we know it is positive.

So, we proved that \(\mathop {\lim }\limits_{x \to c} \frac{{g\left( x \right)}}{{f\left( x \right)}} = 0\) if \(L = 0\).

Case 2 : \(L > 0\)

Let \(\varepsilon > 0\) then because we know \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) there exists a \({\delta _{\,1}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) we have,

\[f\left( x \right) > \frac{{3L}}{{2\varepsilon }} > 0\]

Next, because we know \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) there exists a \({\delta _{\,2}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have,

\[0 < \left| {g\left( x \right) - L} \right| < \frac{L}{2}\hspace{0.25in}\, \to \,\,\,\,\,\,\, - \frac{L}{2} < g\left( x \right) - L < \frac{L}{2}\hspace{0.25in}\, \to \hspace{0.25in}\,\frac{L}{2} < g\left( x \right) < \frac{{3L}}{2}\]

Also, because we are assuming that \(L > 0\) it is safe to assume that for \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have \(g\left( x \right) > 0\).

Now, let \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) and so if \(0 < \left| {x - c} \right| < \delta \) we know from the above statements that we will have both,

\[f\left( x \right) > \frac{{3L}}{{2\varepsilon }}\hspace{0.5in}\hspace{0.25in}g\left( x \right) < \frac{{3L}}{2}\]

This gives us,

\[\left| {\frac{{g\left( x \right)}}{{f\left( x \right)}}} \right| = \frac{{g\left( x \right)}}{{f\left( x \right)}} < \frac{{{}^{{3L}}/{}_{2}}}{{\left| {f\left( x \right)} \right|}} < \frac{{{}^{{3L}}/{}_{2}}}{{{}^{{3L}}/{}_{{2\varepsilon }}}} = \varepsilon \]

In the second step we could remove the absolute value bars because we know or can safely assume (as noted above) that both functions were positive.

So, we proved that \(\mathop {\lim }\limits_{x \to c} \frac{{g\left( x \right)}}{{f\left( x \right)}} = 0\) if \(L > 0\).

Case 3 : \(L < 0\).

Let \(\varepsilon > 0\) then because we know \(\mathop {\lim }\limits_{x \to c} f\left( x \right) = \infty \) there exists a \({\delta _{\,1}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,1}}\) we have,

\[f\left( x \right) > \frac{{ - 3L}}{{2\varepsilon }} > 0\]

Next, because we know \(\mathop {\lim }\limits_{x \to c} g\left( x \right) = L\) there exists a \({\delta _{\,2}} > 0\) such that if \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have,

\[0 < \left| {g\left( x \right) - L} \right| < - \frac{L}{2}\hspace{0.25in}\, \to \,\,\,\,\,\,\,\frac{L}{2} < g\left( x \right) - L < - \frac{L}{2}\hspace{0.25in}\, \to \hspace{0.25in}\,\frac{{3L}}{2} < g\left( x \right) < \frac{L}{2}\]

Next, multiply this be a negative sign to get,

\[ - \frac{L}{2} < - g\left( x \right) < - \frac{{3L}}{2}\]

Also, because we are assuming that \(L < 0\) it is safe to assume that for \(0 < \left| {x - c} \right| < {\delta _{\,2}}\) we have \(g\left( x \right) < 0\).

Now, let \(\delta = \min \left\{ {{\delta _{\,1}},{\delta _{\,2}}} \right\}\) and so if \(0 < \left| {x - c} \right| < \delta \) we know from the above statements that we will have both,

\[f\left( x \right) > \frac{{ - 3L}}{{2\varepsilon }}\hspace{0.5in}\hspace{0.25in} - g\left( x \right) < \frac{{ - 3L}}{2}\]

This gives us,

\[\left| {\frac{{g\left( x \right)}}{{f\left( x \right)}}} \right| = \frac{{ - g\left( x \right)}}{{f\left( x \right)}} < \frac{{{}^{{ - 3L}}/{}_{2}}}{{\left| {f\left( x \right)} \right|}} < \frac{{{}^{{ - 3L}}/{}_{2}}}{{{}^{{ - 3L}}/{}_{{2\varepsilon }}}} = \varepsilon \]

In the second step we could remove the absolute value bars by adding in the negative because we know that \(f\left( x \right) > 0\) and can safely assume that \(g\left( x \right) < 0\) (as noted above).

So, we proved that \(\mathop {\lim }\limits_{x \to c} \frac{{g\left( x \right)}}{{f\left( x \right)}} = 0\) if \(L < 0\).

This is actually a fairly simple proof but we’ll need to do three separate cases.

Case 1 : Assume that \(c > 0\). Next, let \(\varepsilon > 0\) and define

\[M = \sqrt[r]{{\frac{c}{\varepsilon }}}\]

Note that because \(c\) and \(\varepsilon \) are both positive we know that this root will exist. Now, assume that we have

\[x > M = \sqrt[r]{{\frac{c}{\varepsilon }}}\]

Given this assumption we have,

\[\begin{align*}x > \sqrt[r]{{\frac{c}{\varepsilon }}} & & \\ {x^r} & > \frac{c}{\varepsilon }& \hspace{0.5in} & {\mbox{get rid of the root}}\\ \frac{c}{{{x^r}}} & < \varepsilon & \hspace{0.5in} & {\mbox{rearrange things a little}}\\ \left| {\frac{c}{{{x^r}}} - 0} \right| & < \varepsilon & \hspace{0.5in} & {\mbox{everything is positive so we can add absolute value bars}}\end{align*}\]

So, provided \(c > 0\) we’ve proven that \(\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = 0\).

Case 2 : Assume that \(c = 0\). Here all we need to do is the following,

\[\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = \mathop {\lim }\limits_{x \to \infty } \frac{0}{{{x^r}}} = \mathop {\lim }\limits_{x \to \infty } 0 = 0\]

Case 3 : Finally, assume that \(c < 0\). In this case we can then write \(c = - k\) where \(k > 0\). Then using Case 1 and the fact that we can factor constants out of a limit we get,

\[\mathop {\lim }\limits_{x \to \infty } \frac{c}{{{x^r}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - k}}{{{x^r}}} = - \mathop {\lim }\limits_{x \to \infty } \frac{k}{{{x^r}}} = - 0 = 0\]

This is very similar to the proof of 1 so we’ll just do the first case (as it’s the hardest) and leave the other two cases up to you to prove.

Case 1 : Assume that \(c > 0\). Next, let \(\varepsilon > 0\) and define

\[N = - \sqrt[r]{{\frac{c}{\varepsilon }}}\]

Note that because \(c\) and \(\varepsilon \) are both positive we know that this root will exist. Now, assume that we have

\[x < N = - \sqrt[r]{{\frac{c}{\varepsilon }}}\]

Note that this assumption also tells us that \(x\) will be negative. Give this assumption we have,

\[\begin{align*}x & < - \sqrt[r]{{\frac{c}{\varepsilon }}} & & \\ \left| x \right| & > \left| {\sqrt[r]{{\frac{c}{\varepsilon }}}} \right| & \hspace{0.25in} & {\mbox{take absolute value of both sides}}\\ \left| {{x^r}} \right| & > \left| {\frac{c}{\varepsilon }} \right| & \hspace{0.25in} & {\mbox{get rid of the root}}\\ \left| {\frac{c}{{{x^r}}}} \right| & < \left| \varepsilon \right| = \varepsilon & \hspace{0.25in} & {\mbox{rearrange things a little and use the fact that }}\varepsilon > 0\\ \left| {\frac{c}{{{x^r}}} - 0} \right| & < \varepsilon & \hspace{0.25in} & {\mbox{rewrite things a little}}\end{align*}\]

So, provided \(c > 0\) we’ve proven that \(\mathop {\lim }\limits_{x \to - \infty } \frac{c}{{{x^r}}} = 0\). Note that the main difference here is that we need to take the absolute value first to deal with the minus sign. Because both sides are negative we know that when we take the absolute value of both sides the direction of the inequality will have to switch as well.

Case 2, Case 3 : As noted above these are identical to the proof of the corresponding cases in the first proof and so are omitted here.

We’re going to prove this in an identical fashion to the problems that we worked in this section involving polynomials. We’ll first factor out \({a_n}{x^n}\) from the polynomial and then make a giant use of Fact 1 (which we just proved above) and the basic properties of limits.

\[\begin{align*}\mathop {\lim }\limits_{x \to \infty } p\left( x \right) & = \mathop {\lim }\limits_{x \to \infty } \left( {{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + \cdots + {a_1}x + {a_0}} \right)\\ & = \mathop {\lim }\limits_{x \to \infty } \left[ {{a_n}{x^n}\left( {1 + \frac{{{a_{n - 1}}}}{{{a_n}x}} + \cdots + \frac{{{a_1}}}{{{a_n}{x^{n - 1}}}} + \frac{{{a_0}}}{{{a_n}{x^n}}}} \right)} \right]\end{align*}\]

Now, clearly the limit of the second term is one and the limit of the first term will be either \(\infty \) or \( - \infty \) depending upon the sign of \({a_n}\). Therefore by the Facts from the Infinite Limits section we can see that the limit of the whole polynomial will be the same as the limit of the first term or,

\[\mathop {\lim }\limits_{x \to \infty } p\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {{a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + \cdots + {a_1}x + {a_0}} \right) = \mathop {\lim }\limits_{x \to \infty } {a_n}{x^n}\]

The proof of this part is literally identical to the proof of the first part, with the exception that all \(\infty \)’s are changed to \( - \,\infty \), and so is omitted here.

Let \(\varepsilon > 0\) then we need to show that there is a \(\delta > 0\) such that,

\[\left| {f\left( {g\left( x \right)} \right) - f\left( b \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

Let’s start with the fact that \(f\left( x \right)\) is continuous at \(x = b\). Recall that this means that \(\mathop {\lim }\limits_{x \to b} f\left( x \right) = f\left( b \right)\) and so there must be a \({\delta _{\,1}} > 0\) so that,

\[\left| {f\left( x \right) - f\left( b \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - b} \right| < {\delta _{\,1}}\]

Now, let’s recall that \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = b\). This means that there must be a \(\delta > 0\) so that,

\[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta \]

But all this means that we’re done.

Let’s summarize up. First assume that \(0 < \left| {x - a} \right| < \delta \). This then tells us that,

\[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\]

But, we also know that if \(0 < \left| {x - b} \right| < {\delta _{\,1}}\) then we must also have \(\left| {f\left( x \right) - f\left( b \right)} \right| < \varepsilon \). What this is telling us is that if a number is within a distance of \({\delta _{\,1}}\) of \(b\) then we can plug that number into \(f\left( x \right)\) and we’ll be within a distance of \(\varepsilon \) of \(f\left( b \right)\).

So, \[\left| {g\left( x \right) - b} \right| < {\delta _{\,1}}\] is telling us that \(g\left( x \right)\) is within a distance of \({\delta _{\,1}}\) of \(b\) and so if we plug it into \(f\left( x \right)\) we’ll get,

\[\left| {f\left( {g\left( x \right)} \right) - f\left( b \right)} \right| < \varepsilon \]

and this is exactly what we wanted to show.