Test 1A

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Mock Exam (Test 1)

You can try timing yourself for 80 minutes.

Exercise \(\PageIndex{1}\)

Evaluate each of the following limits or explain why they do not exist.

Exercise \(\PageIndex{1.1.1}\)

\(\lim_{h\to 0} \displaystyle \frac{\displaystyle \frac{1}{4+h}-\displaystyle \frac{1}{4}}{h}\)

Answer

\(\frac{-1}{16}\)

Solution

\begin{align*}\lim_{h\to 0} \displaystyle \frac{\displaystyle \frac{1}{4+h}-\displaystyle \frac{1}{4}}{h} &= \lim_{h\to 0} \displaystyle \frac{1}{h} \left(\displaystyle \frac{1}{4+h}-\displaystyle \frac{1}{4} \right)\\&= \lim_{h\to 0} \displaystyle \frac{1}{h}\left(\displaystyle \frac{4-(4+h)}{4(4+h)}\right)\\&= \lim_{h\to 0} \displaystyle \frac{1}{h}\left(\displaystyle \frac{4-4-h}{4(4+h)}\right)\\&= \lim_{h\to 0} \displaystyle \frac{1}{h}\left(\displaystyle \frac{-h}{4(4+h)}\right)\\&= \lim_{h\to 0} \displaystyle \displaystyle \frac{-1}{4(4+h)}\\&= \lim_{h\to 0} \displaystyle \displaystyle \frac{-1}{4(4)}\\&= \displaystyle \displaystyle \frac{-1}{16}\end{align*}

Exercise \(\PageIndex{1.2}\)

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

Answer

\(7\)

Solution

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 \(\PageIndex{1.3}\)

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

Answer

\( \displaystyle \frac{1 }{12}\)

Solution

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

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

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

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

\(=\displaystyle \frac{1 }{2(\sqrt{9}+3)} = \displaystyle \frac{1 }{12}\\(\PageIndex{1.4}\)

Exercise \(\PageIndex{2}\)

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

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

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

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

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

\(\displaystyle \lim_{x\rightarrow \infty} f(x)\)

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

Find any horizontal and vertical asymptotes of this function.

Answer

\(5,5,\infty,-infty,2,2,x=1,y=2\)

Solution

\(\displaystyle\lim_{x\rightarrow 2^+} f(x) =5\) \(\displaystyle \lim_{x\rightarrow 2^-} f(x)=5\) \(\displaystyle \lim_{x\rightarrow 1^+} f(x) =\infty\) \(\displaystyle \lim_{x\rightarrow 1^-} f(x)=-\infty\) \(\displaystyle \lim_{x\rightarrow \infty} f(x)=2\) \(\displaystyle \lim_{x\rightarrow -\infty} f(x)=2\) Horizontal and vertical asymptotes of this function are \(x=1\) and \(y=2\).

 

Exercise \(\PageIndex{3}\)

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 \(\PageIndex{4}\)

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 \(\PageIndex{5}\)

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

\(\frac{3}{50}\)