Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

Learning from Mistakes

There are mistakes in each of the following solutions. Identify the errors, and give the correct answer.

1. Draw the graph of a single function $f$, such that
   • $\underset{x\to 0}{\lim }f\left(x\right)=3$,
   • $\underset{x\to -2}{\lim }f\left(x\right)=\infty$, and
   • $\underset{x\to \infty }{\lim }f\left(x\right)=-1\text{.}$

   Erroneous Solution

   

   Figure 3.37. Erroneous solution to “Learning from Mistakes” Question $1$

   

   Figure 3.38. Graph for “Learning from Mistakes” Question $2\text{.}$

2. Find the following limits for the function shown in Figure 3.38, above:
   1. $\underset{x\to -4^{-}}{\lim }f\left(x\right)\text{.}$
   2. $\underset{x\to 2}{\lim }f\left(x\right)\text{.}$
   3. $\underset{x\to \infty }{\lim }f\left(x\right)\text{.}$
   4. $\underset{x\to -4^{+}}{\lim }f\left(x\right)\text{.}$

   Erroneous Solutions

   1. $\underset{x\to -4^{-}}{\lim }f\left(x\right)=-\infty$
   2. $\underset{x\to 2}{\lim }f\left(x\right)=\infty$
   3. $\underset{x\to \infty }{\lim }f\left(x\right)=2$
   4. $\underset{x\to -4^{+}}{\lim }f\left(x\right)=\infty \text{.}$
3. Evaluate each of the following limits. If a limit does not exist, explain why.
   1. ${\displaystyle \underset{x\to 0^{+}}{\lim }x\sin x+\frac{1}{x}\text{.}}$
   2. ${\displaystyle \underset{x\to 1}{\lim }\frac{x^2+2x-15}{x^2+x-12}\text{.}}$
   3. ${\displaystyle \underset{x\to 0}{\lim }\frac{x^2-1}{x^2+x^4}\text{.}}$
   4. $\underset{x\to \pi ∕2^{-}}{\lim }\tan x+\sec x\text{.}$
   5. ${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{tan\left(3x\right)}{\sin \left(4x\right)}\text{.}}$
   6. ${\displaystyle \underset{x\to 3}{\lim }\frac{\sin \left(x-3\right)}{x^2+x-12}\text{.}}$
   7. ${\displaystyle \underset{x\to \infty }{\lim }x\cos \left(\frac{1}{x}\right)\text{.}}$
   8. ${\displaystyle \underset{x\to \infty }{\lim }\frac{\cos x}{x}\text{.}}$
   9. ${\displaystyle \underset{x\to \infty }{\lim }\frac{x^2+3}{\sqrt{x^2+x-1}}\text{.}}$
   10. ${\displaystyle \underset{x\to -\infty }{\lim }\frac{\sqrt{\left(3x+2\right)\left(x+1\right)}}{3-x}\text{.}}$

   Erroneous Solutions

   1. ${\displaystyle \underset{x\to 0^{+}}{\lim }x\sin x+\frac{1}{x}=\underset{x\to 0^{+}}{lim}x\underset{x\to 0^{+}}{\lim }\sin x+\underset{x\to 0^{+}}{lim}\frac{1}{x}=0-\infty =-\infty \text{.}}$
   2. ${\displaystyle \underset{x\to 1}{\lim }\frac{x^2+2x-15}{x^2+x-12}=\frac{\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x+4\right)}=\frac{x+5}{x+4}=\frac{6}{5}\text{.}}$
   3. ${\displaystyle \underset{x\to 0}{\lim }\frac{x^2-1}{x^2+x^4}=\frac{-1}{0}-\infty \text{.}}$
   4. $\underset{x\to \pi ∕2^{-}}{\lim }\tan x+\sec x=\infty +\infty =\infty \text{.}$
   5. ${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{tan\left(3x\right)}{\sin \left(4x\right)}=\frac{0}{0}\text{;}}$ the limit does not exist because the numerator and denominator get very close to 0.
   6. ${\displaystyle \underset{x\to 3}{\lim }\frac{\sin \left(x-3\right)}{x^2+x-12}=\infty }$ because for $x$ close to 3, ${\displaystyle \frac{1}{x^2+x-12}}$ is very large and $\sin \left(x-3\right)$ oscillates between $-1$ and $1\text{.}$
   7. ${\displaystyle \underset{x\to \infty }{\lim }x\cos \left(\frac{1}{x}\right)=\infty }$ by the Squeeze Theorem because ${\displaystyle -x\le x\cos \left(\frac{1}{x}\right)\le x}$ and ${\displaystyle \underset{x\to \infty }{\lim }\frac{1}{x}=\infty }$
   8. ${\displaystyle \underset{x\to \infty }{\lim }\frac{\cos x}{x}}$ does not exist because $\underset{x\to \infty }{\lim }\cos x$ does not exist.
   9. ${\displaystyle \underset{x\to \infty }{\lim }\frac{x^2+3}{\sqrt{x^2+x-1}}=1}$, because the degree of the numerator is equal to $2$, and the degree of the denominator is equal to $2\text{.}$
   10. $\begin{array}{rl}{\displaystyle \frac{\sqrt{\left(3x+2\right)\left(x+1\right)}}{3-x}}& ={\displaystyle \frac{\sqrt{3x^2+5x+2}}{3-x}}\\ & ={\displaystyle \frac{\sqrt{x^2\left(3+{\displaystyle \frac{5}{x}}+{\displaystyle \frac{2}{x^2}}\right)}}{x\left(3-{\displaystyle \frac{1}{x}}\right)}}\\ & ={\displaystyle \frac{\sqrt[x]{\left(3+{\displaystyle \frac{5}{x}}+{\displaystyle \frac{2}{x^2}}\right)}}{x\left(3-{\displaystyle \frac{1}{x}}\right)}}\\ & ={\displaystyle \frac{\sqrt{\left(3+{\displaystyle \frac{5}{x}}+{\displaystyle \frac{2}{x^2}}\right)}}{\left(3-{\displaystyle \frac{1}{x}}\right)}}\end{array}$
       
       $\begin{array}{rl}\underset{x\to -\infty }{\lim }{\displaystyle \frac{\sqrt{\left(3x+2\right)\left(x+1\right)}}{3-x}}& =\underset{x\to -\infty }{\lim }{\displaystyle \frac{\sqrt{\left(3+{\displaystyle \frac{5}{x}}+{\displaystyle \frac{2}{x^2}}\right)}}{\left(3-{\displaystyle \frac{1}{x}}\right)}}\\ & ={\displaystyle \frac{\sqrt{3}}{-1}}\\ & =-\sqrt{3}\text{.}\end{array}$

4. Find the vertical and horizontal asymptotes of the function

   ${\displaystyle g\left(x\right)=\frac{\sqrt{x^4+4x^3+4x^2}}{3x^2+5x-2}\text{.}}$

   Erroneous Solution

   The function $g$ is not defined at $-2$ and $1∕3$, so $x=-2$ and $x=1∕3$ are vertical asymptotes.

   ${\displaystyle \underset{x\to \infty }{\lim }\frac{\sqrt{x^4+4x^3+4x^2}}{3x^2+5x-2}=\infty \text{,}}$

   because the degree of the numerator is $4$ and that of the denominator is $2$; hence, there are no horizontal asymptotes.