Difference between revisions of "009A Sample Final A, Problem 1"

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(Created page with "<span style="font-size:135%"><font face=Times Roman> 1. Find the following limits:<br>   (a)   <math style="vertical-align: -45%;">\lim_{x\rightarrow0}\frac{\ta...")
 
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. Unfortunately, most (but not all) exam questions require more work.  Many of them will evaluate to an '''indeterminate form''', or something of the form
+
|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,
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac {1}{\infty} = 0,</math>
 +
|-
 +
|and
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow 0^{\pm}}\frac{1}{x}\,=\,\pm \infty.</math>
 +
|-
 +
|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
 
|-
 
|-
 
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{0}{0}</math>&nbsp;&nbsp; or &nbsp;&nbsp;<math>\frac {\pm \infty}{\pm \infty}.</math>
 
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{0}{0}</math>&nbsp;&nbsp; or &nbsp;&nbsp;<math>\frac {\pm \infty}{\pm \infty}.</math>
 
|-
 
|-
−
|<br>In this case, here are several approaches to try:
+
|<br>In this case, there 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 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, <math style="vertical-align: -5%;">x-1</math> can be factored as <math style="vertical-align: -45%;">\left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)</math>&thinsp;, or as <math style="vertical-align: -60%;">\left(\sqrt[3]{x}-1\right)\left(\left(\sqrt[3]{x}\right)^{2}+\sqrt[3]{x}+1\right)</math>&thinsp;, both of which could result in a factor that cancels in our fraction.
+
*We can factor a term creatively.  For example, <math style="vertical-align: -9%;">x-1</math> can be factored as <math style="vertical-align: -48%;">\left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)</math>&thinsp;, or as <math style="vertical-align: -71%;">\left(\sqrt[3]{x}-1\right)\left(\left(\sqrt[3]{x}\right)^{2}+\sqrt[3]{x}+1\right)</math>&thinsp;, both of which could result in a factor that cancels in our fraction.
−
*We can apply '''l'H&ocirc;pital's Rule:''' Suppose <math style="vertical-align: 0%;">c</math> is contained in some interval <math style="vertical-align: 0%;">I</math>. If <math style="vertical-align: -50%;">\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty</math>&thinsp; and <math style="vertical-align: -75%;">\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> &nbsp; exists, and <math style="vertical-align: -20%;">g'(x)\neq 0</math>&thinsp; for all <math style="vertical-align: -20%;">x\neq c</math> in <math style="vertical-align: 0%;">I</math>, then <math style="vertical-align: -73%;">\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.  
+
*We can apply '''l'H&ocirc;pital's Rule:''' Suppose <math style="vertical-align: 0%;">c</math> is contained in some interval <math style="vertical-align: 0%;">I</math>. If <math style="vertical-align: -60%;">\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty</math>&thinsp; and <math style="vertical-align: -85%;">\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> &nbsp; exists, and <math style="vertical-align: -24%;">g'(x)\neq 0</math>&thinsp; for all <math style="vertical-align: -23%;">x\neq c</math>&thinsp; in <math style="vertical-align: 0%;">I</math>, then <math style="vertical-align: -84%;">\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>.  
 
|-
 
|-
 
|Note that the first requirement in l'H&ocirc;pital's Rule is that the fraction <u>''must''</u> be an indeterminate form.  This should be shown in your answer for any exam question.<br>
 
|Note that the first requirement in l'H&ocirc;pital's Rule is that the fraction <u>''must''</u> be an indeterminate form.  This should be shown in your answer for any exam question.<br>
 
|}
 
|}
 +
 +
&nbsp;'''Solution:'''
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Part (a): &nbsp;
 
!Part (a): &nbsp;
 +
|-
 +
|Note that both the numerator and denominator are continuous functions, and that the limit of each is 0 as <math style="vertical-align: 0%;">x</math> approaches 0.  This is an indeterminate form, and we can apply l'H&ocirc;pital's Rule:
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow0}\frac{\tan(3x)}{x^{3}}\,\,\overset{l'H}{=}\,\,\lim_{x\rightarrow0}\frac{\sec^{2}(3x)\cdot3}{3x^{2}}\,=\,\lim_{x\rightarrow0}\frac{3}{3x^{2}}.</math>
 +
|-
 +
|Now, <math style="vertical-align: 0%;">x^2</math> can only be positive, so our limit can also only be positive.  Thus, the limit is <math style="vertical-align: -0%;">+\infty</math>&thinsp;.<br>
 +
|}
  
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Part (b): &nbsp;
 +
|-
 +
|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
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>-\,\frac{\sqrt{\frac{1}{x^{6}}}}{\frac{1}{x^{3}}}\,=\,1,</math>
 +
|-
 +
|since for negative values of <math style="vertical-align: 0%;">x</math>,
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\sqrt{\frac{1}{x^{6}}}\,=\,\left|\frac{1}{x^{3}}\right|\,=\,-\,\frac{1}{x^{3}}.</math>
 +
|-
 +
|This means that
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\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)</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,\lim_{x\rightarrow-\infty}-\,\frac{\sqrt{1+\frac{6x^{2}}{x^{6}}+\frac{2}{x^{6}}}}{1+\frac{x}{x^{3}}+\frac{1}{x}}</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,\lim_{x\rightarrow-\infty}-\,\frac{\sqrt{1+\frac{6}{x^{4}}+\frac{2}{x^{6}}}}{1+\frac{1}{x^{2}}+\frac{1}{x}}</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,-\,\frac{\sqrt{1+0+0}}{1+0+0}</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,-1.</math><br>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Part (c): &nbsp;
 +
|-
 +
|Here, both the numerator and denominator go to zero as <math style="vertical-align: 0%;">x</math> goes to 3, so we have an indeterminate form.  We can choose to either apply l'H&ocirc;pital's Rule, or use the conjugate of the denominator.  Using the conjugate, we find
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\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}}</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,\lim_{x\rightarrow3}\frac{(x-3)(\sqrt{x+1}+2)}{x+1-4}</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,\lim_{x\rightarrow3}\frac{(x-3)(\sqrt{x+1}+2)}{x-3}</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,\lim_{x\rightarrow3}\sqrt{x+1}+2</math>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <math>=\,4.</math>
 +
|-
 +
|Alternatively, we can apply l'H&ocirc;pital's Rule:
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\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.</math><br>
 +
|}
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Part (d): &nbsp;
 +
|-
 +
|This problem is meant to confuse you.  It <u>''looks''</u> like you should multiply by a conjugate, but instead you can just plug in and evaluate:
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow3}\frac{x-1}{\sqrt{x+1}-1}\,=\,\frac{3-1}{\sqrt{3+1}-1}\,=\,2.</math><br>
 +
|}
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Part (e): &nbsp;
 +
|-
 +
|Here, we again multiply by a fraction equal to one, noticing that for all <math style="vertical-align: 0%;">x</math>,
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}\,=\,1.</math>
 +
|-
 +
|This means that
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{x\rightarrow\infty}\frac{5x^{2}-2x+3}{1-3x^{2}}\,=\,\lim_{x\rightarrow\infty}\frac{5x^{2}-2x+3}{1-3x^{2}}\cdot\frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}\,=\,\lim_{x\rightarrow\infty}\frac{5-\frac{2}{x}+\frac{3}{x^{2}}}{\frac{1}{x^{2}}-3}\,=\,\frac{5-0+0}{0-3}\,=\,-\,\frac{5}{3}.</math><br>
 
|}
 
|}
  
  
 
[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 18:21, 27 March 2015

1. Find the following limits:
   (a)   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\rightarrow0}\frac{\tan(3x)}{x^{3}}.}

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

   (c)   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}.}

   (d)   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-1}{\sqrt{x+1}-1}.}

   (e)  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\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,
     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 {1}{\infty} = 0,}
and
     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\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

     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{0}{0}}    or   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 {\pm \infty}{\pm \infty}.}

In this case, there 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, 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-1} can be factored 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 \left(\sqrt{x}-1\right) \left(\sqrt{x}+1\right)}  , or 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 \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 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 c} is contained in some interval 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 I} . If 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 \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty}   and 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\to c}\frac{f'(x)}{g'(x)}}   exists, and 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 g'(x)\neq 0}   for all 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\neq c}   in 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 I} , then 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\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.

 Solution:

Part (a):  
Note that both the numerator and denominator are continuous functions, and that the limit of each is 0 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} approaches 0. This is an indeterminate form, and 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\rightarrow0}\frac{\tan(3x)}{x^{3}}\,\,\overset{l'H}{=}\,\,\lim_{x\rightarrow0}\frac{\sec^{2}(3x)\cdot3}{3x^{2}}\,=\,\lim_{x\rightarrow0}\frac{3}{3x^{2}}.}
Now, 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^2} can only be positive, so our limit can also only be positive. Thus, the limit is 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 +\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
     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{\frac{1}{x^{6}}}}{\frac{1}{x^{3}}}\,=\,1,}
since for negative values of 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} ,
     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 \sqrt{\frac{1}{x^{6}}}\,=\,\left|\frac{1}{x^{3}}\right|\,=\,-\,\frac{1}{x^{3}}.}
This means that
     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\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)}
                                                     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\rightarrow-\infty}-\,\frac{\sqrt{1+\frac{6x^{2}}{x^{6}}+\frac{2}{x^{6}}}}{1+\frac{x}{x^{3}}+\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 =\,\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.}
Part (d):  
This problem is meant to confuse you. It looks like you should multiply by a conjugate, but instead you can just plug in and evaluate:
     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-1}{\sqrt{x+1}-1}\,=\,\frac{3-1}{\sqrt{3+1}-1}\,=\,2.}
Part (e):  
Here, we again multiply by a fraction equal to one, noticing that for all 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} ,
     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{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}\,=\,1.}
This means that
     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\rightarrow\infty}\frac{5x^{2}-2x+3}{1-3x^{2}}\,=\,\lim_{x\rightarrow\infty}\frac{5x^{2}-2x+3}{1-3x^{2}}\cdot\frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}\,=\,\lim_{x\rightarrow\infty}\frac{5-\frac{2}{x}+\frac{3}{x^{2}}}{\frac{1}{x^{2}}-3}\,=\,\frac{5-0+0}{0-3}\,=\,-\,\frac{5}{3}.}


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