Section 4.10 : L'Hospital's Rule and Indeterminate Forms
11. Use L’Hospital’s Rule to evaluate \(\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}\).
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Start SolutionThe first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on certain classes of rational functions and this is clearly not a rational function.
Note however that it is in the following indeterminate form,
\[{\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}{\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} \to {\infty ^0}\]and as we discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L’Hospital’s Rule.
Show Step 2First, let’s define,
\[z = {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}\]and take the log of both sides. We’ll also do a little simplification.
\[\ln z = \ln \left( {{{\left[ {{{\bf{e}}^x} + x} \right]}^{{}^{1}/{}_{x}}}} \right) = \frac{1}{x}\ln \left[ {{{\bf{e}}^x} + x} \right] = \frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}\] Show Step 3We can now take the limit as \(x \to \infty \) of this.
\[\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right]\]Before we proceed let’s notice that we have the following,
\[{\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x} \to = \frac{\infty }{\infty }\]and we have a limit that we can use L’Hospital’s Rule on.
Show Step 4Applying L’Hospital’s Rule gives,
\[\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{}^{{{{\bf{e}}^x} + 1}}/{}_{{{{\bf{e}}^x} + x}}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}}\] Show Step 5We now have a limit that behaves like,
\[{\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} \to \frac{\infty }{\infty }\]and so we can use L’Hospital’s Rule on this as well. Doing this gives,
\[\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x} + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) = 1\]Notice that we did have to use L’Hospital’s Rule twice here and we also made sure to do some simplification so we could actually take the limit.
Show Step 6Now all we need to do is recall that,
\[z = {{\bf{e}}^{\ln z}}\]This in turn means that we can do the original limit as follows,
\[\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} = \mathop {\lim }\limits_{x \to \infty } z = \mathop {\lim }\limits_{x \to \infty } {{\bf{e}}^{\ln z}} = {{\bf{e}}^{\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right]}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\bf{e}}}\]