Calculus I - Infinite Limits

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Section 2.6 : Infinite Limits

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3. For $\displaystyle g\left( z \right) = \frac{{z + 3}}{{{{\left( {z + 1} \right)}^2}}}$ evaluate,

$\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} g\left( z \right)$
$\mathop {\lim }\limits_{z \to \, - {1^{\, + }}} g\left( z \right)$
$\mathop {\lim }\limits_{z \to \, - 1} g\left( z \right)$

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$\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} g\left( z \right)$ Show Solution

Let’s start off by acknowledging that for $z \to - {1^ - }$ we know $z < - 1$ .

For the numerator we can see that, in the limit, we will get 2.

Now let’s take care of the denominator. In the limit, we will have,

$z + 1 \to {0^ - }$

and upon squaring the $z + 1$ we see that, in the limit, we will have,

${\left( {z + 1} \right)^2} \to {0^ + }$

So, in the limit, the numerator is approaching a positive number and the denominator is an increasingly small positive number. The quotient must then be an increasing large positive number or,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} \frac{{z + 3}}{{{{\left( {z + 1} \right)}^2}}} = \infty }}$

Note that this also means that there is a vertical asymptote at $z = - 1$ .

$\mathop {\lim }\limits_{z \to \, - {1^{\, + }}} g\left( z \right)$ Show Solution

Let’s start off by acknowledging that for $z \to - {1^ + }$ we know $z > - 1$ .

For the numerator we can see that, in the limit, we will get 2.

Now let’s take care of the denominator. In the limit, we will have,

$z + 1 \to {0^ + }$

and upon squaring the $z + 1$ we see that, in the limit, we will have,

${\left( {z + 1} \right)^2} \to {0^ + }$

So, in the limit, the numerator is approaching a positive number and the denominator is an increasingly small positive number. The quotient must then be an increasing large positive number or,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{z \to \, - {1^{\, + }}} \frac{{z + 3}}{{{{\left( {z + 1} \right)}^2}}} = \infty }}$

Note that this also means that there is a vertical asymptote at $z = - 1$ , which we already knew from the first part.

$\mathop {\lim }\limits_{z \to \, - 1} g\left( z \right)$ Show Solution

In this case we can see from the first two parts that,

$\mathop {\lim }\limits_{z \to \, - {1^{\, - }}} g\left( z \right) = \mathop {\lim }\limits_{z \to \, - {1^{\, + }}} g\left( z \right) = \infty$

and so, from our basic limit properties we can see that,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{z \to \, - 1} g\left( z \right) = \infty }}$

For the sake of completeness and to verify the answers for this problem here is a quick sketch of the function.