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

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