009A Sample Final A, Problem 1

1. Find the following limits:
   (a)   $\lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}.$

   (b) $\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}.$

   (c)   $\lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}.$

   (d)   $\lim _{x\rightarrow 3}{\frac {x-1}{{\sqrt {x+1}}-1}}.$

   (e)  $\lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}.$

Foundations:
When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. In addition to this, we must consider that as a limit,
${\frac {1}{\infty }}=0,$
and
$\lim _{x\rightarrow 0^{\pm }}{\frac {1}{x}}\,=\,\pm \infty .$
In the latter case, the sign matters. Unfortunately, most (but not all) exam questions require more work. Many of them will evaluate to an indeterminate form, or something of the form
${\frac {0}{0}}$   or   ${\frac {\pm \infty }{\pm \infty }}.$
In this case, here are several approaches to try:
We can multiply the numerator and denominator by the conjugate of the denominator. This frequently results in a term that cancels, allowing us to then just plug in our limit value. We can factor a term creatively. For example, $x-1$ can be factored as $\left({\sqrt {x}}-1\right)\left({\sqrt {x}}+1\right)$ , or as $\left({\sqrt[{3}]{x}}-1\right)\left(\left({\sqrt[{3}]{x}}\right)^{2}+{\sqrt[{3}]{x}}+1\right)$ , both of which could result in a factor that cancels in our fraction. We can apply l'Hôpital's Rule: Suppose $c$ is contained in some interval $I$. If $\lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty$  and $\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$   exists, and $g'(x)\neq 0$  for all $x\neq c$  in $I$, then $\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$.
Note that the first requirement in l'Hôpital's Rule is that the fraction must be an indeterminate form. This should be shown in your answer for any exam question.

Part (a):
Note that both the numerator and denominator are continuous functions, and that the limit of each is 0 as $x$ approaches 0. This is an indeterminate form, and we can apply l'Hôpital's Rule:
$\lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}\,\,{\overset {l'H}{=}}\,\,\lim _{x\rightarrow 0}{\frac {\sec ^{2}(3x)\cdot 3}{3x^{2}}}\,=\,\lim _{x\rightarrow 0}{\frac {3}{3x^{2}}}.$
Now, $x^{2}$ can only be positive, so our limit can also only be positive. Thus, the limit is $+\infty$ .

Part (b):
In the case of limits at infinity, we can apply one other method. We can multiply our original argument by a fraction equal to one, and then can evaluate each term separately. Since we only need to consider values which are negative, we have that
$-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\,=\,1,$
since for negative values of $x$,
${\sqrt {\frac {1}{x^{6}}}}\,=\,\left|{\frac {1}{x^{3}}}\right|\,=\,-\,{\frac {1}{x^{3}}}.$
This means that
$\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\,=\,\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\cdot \left(-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\right)$
$=\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6x^{2}}{x^{6}}}+{\frac {2}{x^{6}}}}}{1+{\frac {x}{x^{3}}}+{\frac {1}{x}}}}$
$=\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6}{x^{4}}}+{\frac {2}{x^{6}}}}}{1+{\frac {1}{x^{2}}}+{\frac {1}{x}}}}$
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,-\,\frac{\sqrt{1+0+0}}{1+0+0}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,-1.}

Part (c):
Here, both the numerator and denominator go to zero as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} goes to 3, so we have an indeterminate form. We can choose to either apply l'Hôpital's Rule, or use the conjugate of the denominator. Using the conjugate, we find
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow3}\frac{x-3}{\sqrt{x+1}-2}\cdot\frac{\sqrt{x+1}+2}{\sqrt{x+1}+2}\,=\,\lim_{x\rightarrow3}\frac{(x-3)(\sqrt{x+1}+2)}{\left(\sqrt{x+1}\right)^{2}-\left(2\right)^{2}}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\lim_{x\rightarrow3}\frac{(x-3)(\sqrt{x+1}+2)}{x+1-4}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\lim_{x\rightarrow3}\frac{(x-3)(\sqrt{x+1}+2)}{x-3}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\lim_{x\rightarrow3}\sqrt{x+1}+2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,4.}
Alternatively, we can apply l'Hôpital's Rule:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow3}\frac{x-3}{\sqrt{x+1}-2}\,\,\overset{l'H}{=}\,\,\lim_{x\rightarrow3}\frac{1}{{\displaystyle \frac{1}{2}\cdot\frac{1}{\sqrt{x+1}}}}\,=\,\frac{1}{\frac{1}{2}\cdot\frac{1}{2}}\,=\,4.}

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