Test 1

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Mock Exam (Test 1)

You can try timing yourself for 60 minutes.

Exercise $\PageIndex{1}$

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

Exercise $\PageIndex{1.1}$

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

Answer

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 $\PageIndex{1.2}$

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

Answer

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 $\PageIndex{1.3}$

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

Answer

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 $\PageIndex{1.4}$

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

Answer

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

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

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

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

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

Exercise $\PageIndex{2}$

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

 

Answer

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 $[0,1]$

Answer

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?

Answer

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.

Answer

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

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

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

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

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

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

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

Exercise $\PageIndex{6}$

a) Use the definition of the derivative to find $f'(x)$ for

$$\displaystyle{f(x)=\frac{1}{3x+1}}$$

Hint:

The definition of the derivative $f'(x)= \lim_{h \rightarrow 0} \displaystyle\frac{f(x+h)-f(x)}{h}$.

Answer

$\displaystyle{f(x)=\frac{-3}{(3x+1)^2}}$.

Solution:

$f'(x)= \lim_{h \rightarrow 0} \displaystyle\frac{f(x+h)-f(x)}{h}$.

b) Find the slope of the tangent line to the graph at the point $x=1.$

Answer

$\displaystyle{f(x)=\frac{-3}{16}}$.

Exercise $\PageIndex{7}$

An object has swimmed the distance $s(t)= \frac{t^2}{t^2+1}$ meters in $t$ seconds. Determine its velocity when $t=3$.

Answer

$\frac{3}{50}$