Elementary Calculus, First Midterm, Copy 1
Second Midterm, Copy 1

Elementary Calculus, First Midterm, Copy 2

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

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


Solution: The notion

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

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


2. Does the limit

(3)   \begin{equation*}\lim _{x \rightarrow 5}\frac{\left \vert x -5\right \vert }{x -5} \end{equation*}



Solution: We know \left \vert x -5\right \vert =x -5 when x -5 \geq 0 and \left \vert x -5\right \vert = -\left (x -5\right ) when x -5 <0. We write

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


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

Now since

(6)   \begin{equation*}\lim _{x \rightarrow 5^{ -}}\frac{\left \vert x -5\right \vert }{x -5} \neq \lim _{x \rightarrow 5^{ +}}\frac{\left \vert x -5\right \vert }{x -5} \end{equation*}

we can conclude

(7)   \begin{equation*}\lim _{x \rightarrow 5}\frac{\left \vert x -5\right \vert }{x -5} \end{equation*}

does not exist.


3. Compute the limit

(8)   \begin{equation*}\lim _{x \rightarrow 2}\frac{x^{3} -8}{2 -x} \end{equation*}


Solution: We start by factoring and then simplifying

(9)   \begin{align*}\lim _{x \rightarrow 2}\frac{x^{3} -8}{2 -x} =\lim _{x \rightarrow 2}\frac{\left (x -2\right )\left (x^{2} +2x +4\right )}{ -\left (x -2\right )} \\ =\lim _{x \rightarrow 2}\frac{x^{2} +2x +4}{ -1} \\ = -12\end{align*}


4. Is the function

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

continuous at x =4 ?


Solution: We note that

(11)   \begin{equation*}f\left (4\right ) =\frac{1}{4^{2} -7\left (4\right ) +10} = -\frac{1}{2} \end{equation*}


(12)   \begin{equation*}\lim _{x \rightarrow 4}f\left (x\right ) =\lim _{x \rightarrow 4}\frac{1}{x^{2} -7x +10} = -\frac{1}{2} \end{equation*}

It is easy to see that

(13)   \begin{equation*}\lim _{x \rightarrow 4}f\left (x\right ) =f\left (4\right ) \end{equation*}

and therefore the function f is continuous at x =4.


5. Find an equation of the tangent line to the graph of f\left (x\right ) =4 -x^{2} at \left ( -1 ,3\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 ,3\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

(14)   \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{4 -\left (x +h\right )^{2} -\left (4 -x^{2}\right )}{h} \\ =\lim _{h \rightarrow 0}\frac{4 -x^{2} -2xh -h^{2} -4 +x^{2}}{h} \\ =\lim _{h \rightarrow 0}\frac{ -2xh -h^{2}}{h} \\ =\lim _{h \rightarrow 0}\left ( -2x -h\right ) \\ = -2x\end{align*}

Therefore the slope is found to be

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

and an equation of the tangent line is

(16)   \begin{equation*}y -3 =2\left (x -\left ( -1\right )\right ) \end{equation*}


6. Find the limit

(17)   \begin{equation*}\lim _{x \rightarrow -2}\frac{1}{x +2} \end{equation*}


Solution: We immediately notice that denominator approaches zero as x approches -2. Since the sign of denominator will be different as x approaches from right and left, we obtain

(18)   \begin{align*}\lim _{x \rightarrow -2^{ -}}\frac{1}{x +2} = -\infty \\ \lim _{x \rightarrow -2^{ +}}\frac{1}{x +2} =\infty \end{align*}

The conclusion is that

(19)   \begin{equation*}\lim _{x \rightarrow -2}\frac{1}{x +2} \end{equation*}

does not exist.


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