Exam 1 - Limits and Derivatives

Part I

(1) Calculate the following limits (if they exist):

(a)

$$\lim_{x \to -2} \frac{x^2+7x+12}{x^2-7x+10}$$

Direct Substitution:

$$=\frac{4-14+12}{4+14+10}=\frac{2}{28}= \frac{1}{14}.$$

(b)

$$\lim_{x \to 0} \frac{\tan x}{x}$$

Direct substitution yields $\frac{0}{0}$, so further analysis is required.

$$=\lim_{x \to 0} \frac{\tan x}{x}= \lim_{x \to 0} \frac{\sin x... ... x}{x} \lim_{x \to 0} \frac{1}{\cos x} =(1)(\frac{1}{\cos 0})=1$$

(c)

$$\lim_{x \to 0} \frac{\sqrt{7+x}-\sqrt{7}}{x}$$

Direct substitution yields $\frac{0}{0}$, so further analysis is required.

$$\lim_{x \to 0} \frac{\sqrt{7+x}-\sqrt{7}}{x} = \lim_{x \to 0}... ...sqrt{7}}{x} (\frac{\sqrt{7+x}+\sqrt{7}} {\sqrt{7+x}+\sqrt{7}})$$

$$= \lim_{x \to 0} \frac{{7+x}-{7}}{x(\sqrt{7+x}+\sqrt{7})} = \... ...\sqrt{7+x}+\sqrt{7}} = \frac{1}{2\sqrt{7}}= \frac{\sqrt{7}}{14}$$

(2) Find the derivatives of the following functions:

(a) $f(x) = (x^2+3)\sin x$

Product Rule: $2x\sin x + (x^2+3)\cos x$

(b) $g(x) = x^3+\sqrt{x}+\frac{1}{x}$

$3x^2+ \frac{1}{2\sqrt{x}} - \frac{1}{x^2}$.

(c) h(x) = (x³+3x²+7x+4)(2x⁵+3x²+2x+1)

Product Rule: (3x²+6x+7)(2x⁵+3x²+2x+1) + (x³+3x²+7x+4)(10x⁴+6x+2)

Part 2

(3) Consider

$$g(x)= \left\{ \begin{array}{ll} x^2-1 & x \leq -1 \\ \frac{1}{x} + 2 & x > -1 \\ \end{array} \right.$$

(a) Where is g(x) not continuous?

There is a jump discontinuity at x=-1 because the right and left limits are different. Also, the function is unbounded at x=0, so this is also a discontinuity.

(b) Identify each discontinuity as removable or nonremovable.

Both x = 0 and x = -1 are nonremovable.

(c) Where are the vertical asymptotes?

x = 0

(4) Calculate the derivative of f(x)= 3x²+x+1 using the definition of the derivative.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{[3{(x+h)}^2 +(x+h)+1] - (3x^2+x+1)}{h}$$

$$= \lim_{h \to 0} \frac{6xh + 3h^2 +h}{h} = \lim_{h \to 0} {6x+3h+1} = 6x+1$$

(5) Find the equation of the tangent line to the curve $y=2\cos x + 3\sin x$ at the point (0,2).

The derivative is $y' = -2\sin x + 3 \cos x$, so the slope at (0,3) is $-2\sin 0 + 3 \cos 0 = 3$. Therefore the equation of the line is: (y-2)=3(x-0).

(6) Do one of the following (one is impossible): Give an example of a function which is continuous but not differentiable. Give an example of a function which is differentiable but not continuous.

f(x) = |x| is continuous but not differentiable at x = 0. There are many examples of this direction, but the other direction is impossible.

(7) Find the velocity and acceleration of a ball which is thrown upwards at a speed of 115 feet per second after 2 seconds. The position equation is given by:

s(t)= -16t²+96t.

Velocity is the first derivative and acceleration the second. So v(t) = s'(t) = -32t+96 and a(t)=v'(t) = -32. Thus v(2) = 32ft/s and a(2) = -32ft/s².

(8) Consider the function f(x) = x³+x². We want to find the slope of the tangent line to the curve of y=f(x) at the point (1,2).

(a) Estimate the slope by taking slopes of secant lines through the points (1.1, (1.1)² +(1.1)³), (1.01, (1.01)²+ (1.01)³), and (1+h,(1+h)² +(1+h)³).

The first two slopes are 5.41 and 5.0401, while the thrid is:

$$m = \frac{{(1+h)}^2 +{(1+h)}^3 - 2}{(1+h)-1} = 5+4h+h^2$$

(b) Find the slope by taking the limit of your third answer from (a) as h approaches 0.

$\lim_{h \to 0} 5 +4h +h^2 = 5$

(c) Verify your answer by finding the derivative of the function the easy way, and evaluating at x = 1.

f'(x) = 3x² + 2x so f'(1) = 3(1) + 2(1) = 5