Power Series
Elementary Calculus, First Midterm, Copy 2

Elementary Calculus, First Midterm, Copy 1

1. Suppose f\left (x\right ) is a given function and a is a real number. State what we mean by

(1)   \begin{equation*}\lim _{x \rightarrow a}f\left (x\right ) =\infty \end{equation*}

 

Solution: The notion

(2)   \begin{equation*}\lim _{x \rightarrow a}f\left (x\right ) =\infty \end{equation*}

means the value of the function f\left (x\right ) can be as large as we like, whenever x is close enough to the number a .

 

2. Does the limit

(3)   \begin{equation*}\lim _{x \rightarrow 0}\frac{\left \vert x\right \vert }{x} \end{equation*}

exist?

 

Solution: We know \left \vert x\right \vert =x when x \geq 0 and \left \vert x\right \vert = -x when x <0. We write

(4)   \begin{equation*}\lim _{x \rightarrow 0^{ -}}\frac{\left \vert x\right \vert }{x} =\lim _{x \rightarrow 0^{ -}}\frac{ -x}{x} = -1 \end{equation*}

Next,

(5)   \begin{equation*}\lim _{x \rightarrow 0^{ +}}\frac{\left \vert x\right \vert }{x} =\lim _{x \rightarrow 0^{ +}}\frac{x}{x} =1 \end{equation*}

Now since

(6)   \begin{equation*}\lim _{x \rightarrow 0^{ -}}\frac{\left \vert x\right \vert }{x} \neq \lim _{x \rightarrow 0^{ +}}\frac{\left \vert x\right \vert }{x} \end{equation*}

we can conclude

(7)   \begin{equation*}\lim _{x \rightarrow 0}\frac{\left \vert x\right \vert }{x} \end{equation*}

does not exist.

 

3. Compute the limit.

(8)   \begin{equation*}\lim _{x \rightarrow 0}\frac{\sqrt{4 +x} -\sqrt{4 -x}}{x} \end{equation*}

 

Solution: We start by simplifying

(9)   \begin{align*}\lim _{x \rightarrow 0}\frac{\sqrt{4 +x} -\sqrt{4 -x}}{x} =\lim _{x \rightarrow 0}\frac{\sqrt{4 +x} -\sqrt{4 -x}}{x}\genfrac{(}{)}{}{}{\sqrt{4 +x} +\sqrt{4 -x}}{\sqrt{4 +x} +\sqrt{4 -x}} \\ =\lim _{x \rightarrow 0}\frac{\left (4 +x\right ) -\left (4 -x\right )}{x\left (\sqrt{4 +x} +\sqrt{4 -x}\right )} \\ =\lim _{x \rightarrow 0}\frac{2x}{x\left (\sqrt{4 +x} +\sqrt{4 -x}\right )} \\ =\lim _{x \rightarrow 0}\frac{2}{\sqrt{4 +x} +\sqrt{4 -x}} \\ =\frac{1}{2}\end{align*}

 

4. Is the function

(10)   \begin{equation*}f\left (x\right ) =\frac{1}{x^{2} -7x +10} \end{equation*}

continuous at x =5 ?

 

Solution: Let us observe that f\left (5\right ) is not defined since

(11)   \begin{equation*}5^{2} -7\left (5\right ) +10 =0 \end{equation*}

Therefore, f\left (x\right ) cannot be continuous at x =5. Note that in order for a function to be continuous at a point, it must be defined at that point.

 

5. Find an equation of the tangent line to the graph of f\left (x\right ) =\frac{2}{x} at \left (1 ,2\right ) .

 

Solution: In order to write the point-slope formula of any line, we need a point on the line and the slope of the line. The point is given \left (1 ,2\right ). To find the slope, we compute the derivative of f since evaluating the derivative at the given point provides the slope of the tangent line at that point. Write

(12)   \begin{align*}f^{ \prime }\left (x\right ) =\lim _{h \rightarrow 0}\frac{f\left (x +h\right ) -f\left (x\right )}{h} \\ =\lim _{h \rightarrow 0}\frac{\frac{2}{x +h} -\frac{2}{x}}{h} \\ =2\lim _{h \rightarrow 0}\frac{\frac{x -\left (x +h\right )}{x(x +h)}}{h} \\ =2\lim _{h \rightarrow 0}\frac{ -1}{x\left (x +h\right )} \\ =\frac{ -2}{x^{2}}\end{align*}

It is now easy to see slope is equal to

(13)   \begin{equation*}f^{ \prime }\left (1\right ) =\frac{ -2}{1^{2}} = -2 \end{equation*}

and so

(14)   \begin{equation*}y -2 = -2\left (x -1\right ) \end{equation*}

is an equation of the tangent line.

 

6. Find the limit

(15)   \begin{equation*}\lim _{x \rightarrow 3}\frac{1}{x -3} \end{equation*}

 

Solution: We immediately notice that denominator approaches zero as x approches 3. Since the sign of denominator will be different as x approaches from right and left, we obtain

(16)   \begin{align*}\lim _{x \rightarrow 3^{ -}}\frac{1}{x -3} = -\infty \\ \lim _{x \rightarrow 3^{ +}}\frac{1}{x -3} =\infty \end{align*}

The conclusion is that

(17)   \begin{equation*}\lim _{x \rightarrow 3}\frac{1}{x -3} \end{equation*}

does not exist.

 

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