$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# Test 1

[ "stage:draft", "article:topic", "authorname:thangarajahp" ]

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

These mock exams are provided to help you prepare for  Term/Final tests.  The best way to use these practice tests is to try the problems as if you were taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand it.

## Mock Exam (Test 1)

You can try timing yourself for 50 minutes.

Exercise $$\PageIndex{1}$$

Calculate the following four limits or explain why they do not exist:

Exercise $$\PageIndex{1.1}$$

$$\displaystyle \lim_{x \to 2} \frac{2x^2-x-6}{x-2}$$

First set $$x=2$$, and we get $$\displaystyle\lim_{x \to 2} \frac{2x^2-x-6}{x-2}= \frac{2(2^2)-2-6}{2-2} \, \left[ \frac{0}{0} \right]$$.

Now, $$\displaystyle \lim_{x \to 2} \frac{2x^2-x-6}{x-2} = \displaystyle\lim_{x \to 2} \frac{(2x+3)(x-2)}{x-2}=\displaystyle\lim_{x \to 2} (2x+3)= 2(2)+3=7.$$

Exercise $$\PageIndex{1.2}$$

$$\lim_{y \to \frac{1}{4} } \mbox{cos}^{-1}(1-2y)$$

Since $$\mbox{cos}^{-1}(y)$$ is continuous on $$[0,\pi]$$ and $$\lim_{y \to \frac{1}{4} }( 1-2y) =1- 2 \left( \frac{1}{4}\right)= \frac{1}{2}$$ exists,

$$\lim_{y \to \frac{1}{4} }\mbox{cos}^{-1}( 1-2y) = \mbox{cos}^{-1} \left( \frac{1}{2}\right) = \frac{\pi}{3}. \ Exercise \(\PageIndex{1.3}$$

$$\displaystyle \lim_{y \to 0} |\sqrt{7y^2+6y+9}|$$

Since absolute value function is continuous and $$\displaystyle\lim_{y \to 0} \sqrt{7y^2+6y+9}$$= 3, $$\displaystyle\lim_{y \to 0} |\sqrt{7y^2+6y+9}|$$=|3|=3. \)

Exercise $$\PageIndex{1.4}$$

$$\displaystyle \lim_{x\rightarrow 9}\frac{\sqrt{x}-3}{2x-18}$$

$$\displaystyle \lim_{x\rightarrow 9}\frac{\sqrt{x}-3}{2x-18} = \frac{\sqrt{9}-3}{18-18} [\frac{0}{0}]$$

$$=\displaystyle \lim_{x\rightarrow 9}\frac{(\sqrt{x}-3)(\sqrt{x}+3) }{(2x-18)(\sqrt{x}+3)}$$

$$=\displaystyle \lim_{x\rightarrow 9}\frac{(x-9) }{2(x-9)(\sqrt{x}+3)}$$

$$=\displaystyle \lim_{x\rightarrow 9}\frac{1 }{2(\sqrt{x}+3)}$$

$$=\displaystyle \frac{1 }{2(\sqrt{9}+3)} = \displaystyle \frac{1 }{12}$$

Exercise $$\PageIndex{2}$$

Determine where $$\displaystyle\frac{\mbox{cos}^{-1}x}{(\mbox{tan}(x)-1)}$$ is continuous.

Since $$\mbox{cos}^{-1}(y)$$ is continuous on $$[-1,1]$$  and $$(\mbox{tan}(x)-1)=0$$ when $$x=\frac{\pi}{4}$$,

$$\displaystyle\frac{\mbox{cos}^{-1}x}{(\mbox{tan}^{-1}x-1)}$$  is continuous on $$[-1,1] \setminus \frac{\pi}{4}$$.  Note that $$\frac{\pi}{4} <1$$.  \

Exercise $$\PageIndex{3}$$

Use the Intermediate Value Theorem to show that the equation

$$\displaystyle 4x^5-6x^3+10x^3-5=0$$ has at least one solution in the interval $$[0,1]$$

Let $$f(x)= 4x^5-6x^3+10x^3-5$$. Then $$f(x)$$ is continuous and $$f(0)=-5$$ and $$f(1)=3$$.

Since $$-5<0<3$$,by the Intermediate Value Theorem there exist  at least one  real number $$k$$ in the interval $$[0,1]$$ such that $$f(k)=0$$

Exercise $$\PageIndex{4}$$

The equation $$\displaystyle Q=12e^{-0.055t}$$ gives the mass $$\displaystyle Q$$ in grams of the radioactive isotope $$\displaystyle\mbox{potassium}^{-42}$$ that will remain from some initial quantity after $$\displaystyle t$$ hours of radio active decay.

a) How many grams are there initially (i.e. at time 0 hours [$$\displaystyle t=0$$])

b) How long will it take to reduce the amount of radioactive isotope $$\displaystyle\mbox{potassium}^{-42}$$ to one third of the original amount?

a)  $$t=0$$, then $$\displaystyle Q=12e^{-0.055(0)} = 12 g$$.

b)  We need to find $$t$$ when $$Q$$ = One third of the original amount  is $$12 (1/3) g$$, Then  $$\displaystyle 4=12e^{-0.055(t)}$$.

Thus $$\displaystyle 1/3=e^{-0.055(t)}$$.

$$t= \frac{-1}{0.055} ln(\frac{1}{3}) = \frac{ln(3)}{0.055}$$.

Exercise $$\PageIndex{5}$$

Consider $$f(x) =\displaystyle \frac{2x+1}{x-1}.$$  Use limits to find any horizontal and vertical asymptotes of this function.

1.  $$\displaystyle\lim_{x\rightarrow 2^+} f(x)$$
2.  $$\displaystyle \lim_{x\rightarrow 2^-} f(x)$$
3. $$\displaystyle \lim_{x\rightarrow 1^+} f(x)$$
4. $$\displaystyle \lim_{x\rightarrow 1^-} f(x)$$
5. $$\displaystyle \lim_{x\rightarrow \infty} f(x)$$
6. $$\displaystyle \lim_{x\rightarrow -\infty} f(x)$$
7. Find any horizontal and vertical asymptotes of this function.

$$\displaystyle\lim_{x\rightarrow 2^+} f(x) =5$$

$$\displaystyle \lim_{x\rightarrow 2^-} f(x)=5$$

$$\displaystyle \lim_{x\rightarrow 1^+} f(x) =\infty$$

$$\displaystyle \lim_{x\rightarrow 1^-} f(x)=-\infty$$

$$\displaystyle \lim_{x\rightarrow \infty} f(x)=2$$

$$\displaystyle \lim_{x\rightarrow -\infty} f(x)=2$$

Horizontal and vertical asymptotes of this function are $$x=1$$ and  $$y=2$$.