Exam 1 Practice - Limits and Derivatives

Part I

(1) Calculate the following limits:

(a)

$$\lim_{x \to 1} \frac{x^2-7x+6}{x^2-2x+1}$$

First try direct substitution, but $\frac{1^2-7+6}{1^2-2+1}=\frac{0}{0}$. Therefore we have to try something else, and considering that we have two rational functions, it seems that factoring might be the way to go. SO

$$\lim_{x \to 1} \frac{x^2-7x+6}{x^2-2x+1}=$$

$$\lim_{x \to 1} \frac{(x-1)(x-6)}{(x-1)(x-1)}$$

$$=\lim_{x \to 1} \frac{(x-6)}{(x-1)}$$

Now we have a problem, since we get $\frac{1}{0}$, and our limit is undefined. A graph shows that we have unbounded behavior (from the right to - $\infty$ and from the left + $\infty$). Saying that the limit is undefined would suffice.

(b)

$$\lim_{x \to 0} \frac{\sin x}{x^2}$$

Again try direct substitution first: again the result is $\frac{0}{0}$, so we have to do something else. This looks quite similar to the limit we saw in class: $\lim_{x \to 0} \frac{\sin x}{x}=1$. We must use this limit:

$$\lim_{x \to 0} \frac{\sin x}{x^2}$$

$$=\lim_{x \to 0} \frac{\sin x}{x}\frac{1}{x}$$

$$=1(\lim_{x \to 0} \frac{1}{x})$$

As in (1) the limit is undefined (unbounded behavior).

(c)

$$\lim_{x \to 3} \frac{\sqrt{1+x}-2}{x-3}$$

For the third straight time try direct substitution, and again we get $\frac{0}{0}$. Now this one has square root in it, so factoring doesn't look good nor can we compare to a known limit. Here looks like a perfect time to use rationalization:

$$\lim_{x \to 3} \frac{\sqrt{1+x}-2}{x-3}$$

$$=\lim_{x \to 3} \frac{\sqrt{1+x}-2}{x-3} \frac{\sqrt{1+x}+2}{\sqrt{1+x}+2}$$

$$=\lim_{x \to 3} \frac{ {(\sqrt{1+x})}^2-2^2}{(x-3)(\sqrt{1+x}+2)}$$

$$=\lim_{x \to 3} \frac{ 1+x-4}{(x-3)(\sqrt{1+x}+2)}$$

$$=\lim_{x \to 3} \frac{(x-3)}{(x-3)(\sqrt{1+x}+2)}$$

$$=\lim_{x \to 3} \frac{1}{(\sqrt{1+x}+2)}$$

$$=\frac{1}{4}$$

(2) Find the derivatives of the following functions:

(a) $f(x) = {(x^3+3\sin x +4)}^8$

Looks like a chain rule problem, with the outer function being raising to the eighth power and the inner $(x^3+3\sin x +4)$. So the derivative is: $f'(x) = 8 {(x^3+3\sin x +4)}^7 (3x^2+3\cos x)$.

(b) $g(x) = x^3\csc x$

This is the product of two functions, so the derivative = the sum of (the first's derivative)*(the second)+ (the first)* (the second's derivative)= $(3x^2)(\csc x) + (x^3)(-\csc x * \cot x) =3x^2\csc x - x^3 \csc x \cot x$.

(c) $h(x) = \frac{x+1}{x+5}$

This one is the quotient of two functions, so it looks like the time and place to use the quotient rule: derivative = $\frac{ (denominator)(derivative of numerator) - (numerator) (derivative of denominator)}{ {(denominator)}^2}$

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

Part 2

(3) Consider $f(x)= \frac{x^3+1}{x^2-5x-6}$.

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

f(x) is not continuous where the denominator is 0, namely at the roots of x²-5x-6, or at -1 and 6.

(b) Identify each discontinuity as removable or nonremovable.

-1 is removable while 6 is not removable. This can be verified by checking that the limit of f(x) does not exist as x approaches 6, but it does exist at the point -1. (You should check this, and show this work on the exam).

(c) Where are the vertical asymptotes?

There is a vertical asymptote of x=6, since the right hand limit at 6 is $\infty$ (also because the left hand limit is $-\infty$-either alone would make there be an asymptote).

(4) Calculate the derivative of $f(x)= \frac{1}{x+1}$ using the definition of the derivative.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

$$=\lim_{h \to 0} \frac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h} =\l... ...frac{\frac{x+1}{(x+1)(x+h+1)}-\frac{(x+h+1)}{(x+1)(x+h+1)}} {h}$$

$$=\lim_{h \to 0} \frac{x+1-x-h-1}{(x+1)(x+h+1)h}= \lim_{h \to 0} \frac{-h}{(x+1)(x+h+1)h}$$

$$=\lim_{h \to 0} \frac{-1}{(x+1)(x+h+1)} = \frac{-1}{(x+1)(x+1)}$$

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

To get the equation of a line we need either: (1) two points or (2) one point and a slope. We have only one point to work with, $(\pi,3)$ so we need to find the slope. The slope of the tangent line at that point on the curve is the derivative at that point:

$$\frac{dy}{dx} = 2 \cos x = 2 \cos (\pi) = -2$$

Therefore the equation of the tangent line is:

$$(y-3)=-2(x-\pi)$$

(6) Find the derivative of y with respect to x for the equation:

y³+3xy-2(x²y)=sin (y).

It would be a great challenge to find an explicit form for the function of y in terms of x, so implicit differentiation is the way to go. First differentiate both sides with respect to x:

$$(3y^2)\frac{dy}{dx}+ [3y+ 3x\frac{dy}{dx}]-2[2xy+x^2\frac{dy}{dx}] =\cos (y) \frac{dy}{dx}$$

The next step is to group the terms with $\frac{dy}{dx}$ on the left, the other terms on the right, so we have:

$$(3y^2)\frac{dy}{dx}+3x\frac{dy}{dx}-2x^2\frac{dy}{dx}-\cos (y) \frac{dy}{dx} =4xy-3y$$

$$\frac{dy}{dx} [3y^2+3x-2x^2- \cos (y)]=4xy-3y$$

$$\frac{dy}{dx}= \frac{4xy-3y}{3y^2+3x-2x^2- \cos (y)}$$

(7) Find the velocity at impact of a bowling ball dropped from a building which is 400 feet high, given that the ball has position equation

s(t)=-16t²+400.

Velocity can be found by taking the derivative of the position function, s(t). v(t)=s'(t)=-32t. But we also need to figure out the time of impact, or when s(t)=0. s(t)=-16t²+400=0 when $t = \pm 5$ but the only one of these which makes physical sense is t=5. Therefore the velocity of impact is v(5)= -160feet/second.