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PAGE 1

Limit at Infinity Calculation

\[ \lim_{x \to \infty} \frac{\sqrt{4x^2 + 2x}}{3x + 1} = \]

Limit at infinity: divide top and bottom by the highest degree of the denominator.

Here, divide by \( x \).

\[ \lim_{x \to \infty} \frac{\frac{\sqrt{4x^2 + 2x}}{x}}{\frac{3x + 1}{x}} = \lim_{x \to \infty} \frac{\frac{\sqrt{4x^2 + 2x}}{\sqrt{x^2}}}{3 + \frac{1}{x}} \]
\[ = \lim_{x \to \infty} \frac{\sqrt{\frac{4x^2 + 2x}{x^2}}}{3 + \frac{1}{x}} = \lim_{x \to \infty} \frac{\sqrt{4 + \frac{2}{x}}}{3 + \frac{1}{x}} \]

As \( x \to \infty \), the terms \( \frac{2}{x} \to 0 \) and \( \frac{1}{x} \to 0 \). Therefore:

\[ = \frac{\sqrt{4}}{3} = \frac{2}{3} \]

Multiple Choice Options

  1. A. \( \infty \)
  2. B. \( \frac{4}{3} \)
  3. C. \( 0 \)
  4. D. \( \frac{2}{3} \) (Correct Answer)
  5. E. does not exist
PAGE 2

Asymptotes of a Function

Choose the right statement which describes ALL the horizontal and vertical asymptotes of the function

\[ f(x) = \frac{e^x + 1}{e^x - 1} \]

Options

  1. A. Horizontal Asymptote(s): \( y = 1, y = -1 \), Vertical Asymptote(s): None
  2. B. Horizontal Asymptote(s): \( y = 1 \), Vertical Asymptote(s): \( x = 1 \)
  3. C. Horizontal Asymptote(s): \( y = 1 \), Vertical Asymptote(s): \( x = 0 \)
  4. D. Horizontal Asymptote(s): \( y = 1, y = -1 \), Vertical Asymptote(s): \( x = 0 \) (Correct Answer)
  5. E. Horizontal Asymptote(s): None, Vertical Asymptote(s): \( x = 0 \)
PAGE 3

Asymptote Analysis

Vertical Asymptote

Vertical asymptote: when denom = 0 while numerator ≠ 0

Here, \( e^x - 1 = 0 \)

\( e^x = 1 \rightarrow \)

x = 0

one vertical asymptote at \( x = 0 \)

Horizontal Asymptote

  • Horizontal asymptote to the right: \( \lim_{x \to \infty} \)
  • Horizontal asymptote to the left: \( \lim_{x \to -\infty} \)

Right: \( \lim_{x \to \infty} \frac{e^x + 1}{e^x - 1} \rightarrow \frac{e^{\text{big \#}} + 1}{e^{\text{big \#}} - 1} \)

Note: \( +1 \) and \( -1 \) are insignificant compared to \( e^x \) as \( x \to \infty \)

\( \lim_{x \to \infty} e^x = \infty \)
\( \lim_{x \to -\infty} e^x = 0 \)

\( = \frac{e^x}{e^x} = 1 \)

Horizontal asymptote (right): y = 1

Left: \( \lim_{x \to -\infty} \frac{e^x + 1}{e^x - 1} \)

As \( x \to -\infty \), \( e^x \to 0 \). Thus:

\( = \frac{0 + 1}{0 - 1} = -1 \)

Horizontal asymptote (left): y = -1
PAGE 4

Continuity Problem

For what value(s) of \( c \) is

\[ f(x) = \begin{cases} -cx + 1, & \text{if } x < 2 \\ 3, & \text{if } x = 2 \\ c^2x^2 + 2, & \text{if } x > 2 \end{cases} \]

continuous from the left at 2.

A. \( -\frac{1}{2}\sqrt{\frac{3}{2}}, \frac{1}{2}\sqrt{\frac{3}{2}} \)

B. 0

C. \( \frac{1}{2} \)

D. 2

E -1

Solution

Continuous from LEFT: \( \lim_{x \to 2^-} f(x) = f(2) \)

\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (-cx + 1) = -2c + 1 \)

from left of 2, so \( x < 2 \), so top branch

We know \( f(2) = 3 \)

Solve:

\( -2c + 1 = 3 \)

\( -2c = 2 \)

c = -1
PAGE 5

Calculus: Derivative at a Point

2. Let \( f(x) = \frac{1}{\sqrt{x}} \). Which of the following equals \( f'(4) \)?

I.
\[ \lim_{h \to 0} \frac{\frac{1}{\sqrt{4+h}} - \frac{1}{\sqrt{4}}}{h} \]
✓
II.
\[ \lim_{x \to 4} \frac{\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{x}}}{x - 4} \]
✗
III.
\[ \frac{-1}{16} \]
✓

Limit Definitions:

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

①

\( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \)

②

Step-by-Step Verification

①: \( f(x) = \frac{1}{\sqrt{x}} \)

\[ f'(4) = \lim_{h \to 0} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{4+h}} - \frac{1}{\sqrt{4}}}{h} \]

②: \( f(x) = \frac{1}{\sqrt{x}} \)

\[ f'(4) = \lim_{x \to 4} \frac{f(x) - f(4)}{x - 4} = \lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{4}}}{x - 4} \]

III: \( f(x) = \frac{1}{\sqrt{x}} = x^{-1/2} \)

\[ f'(x) = -\frac{1}{2}x^{-3/2} = \frac{-1}{2x^{3/2}} \]\[ f'(4) = -\frac{1}{2} \cdot \frac{1}{4^{3/2}} = -\frac{1}{2} \cdot \frac{1}{8} = -\frac{1}{16} \]

Multiple Choice Options

  • A. I. only
  • B. II. only
  • C. III. only
  • D. I. and III. only
  • E. I. and II. and III.
PAGE 6

Limits of Trigonometric Functions

Find the limit.

\[ \lim_{x \to 0} \frac{\sin(4x) \sin(3x)}{x^2} \]
  • A. 0
  • B. 12
  • C. \( \frac{1}{12} \)
  • D. \( \frac{3}{4} \)
  • E. Does not exist.

Special limit

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Solution:

\[ \lim_{x \to 0} \frac{\sin(4x)}{x} \cdot \frac{\sin(3x)}{x} \]

We want the denominator to match the argument of the sine function:

  • For \( \sin(4x) \), we want \( 4x \)
  • For \( \sin(3x) \), we want \( 3x \)

So \( \lim_{x \to 0} \frac{\sin(4x)}{4x} = 1 \)

\[ = \lim_{x \to 0} \underbrace{\frac{\sin(4x)}{4x}}_{1} \cdot \underbrace{\frac{\sin(3x)}{3x}}_{1} \cdot \frac{4 \cdot 3}{1} = 12 \]
PAGE 7

Calculus Problem: Derivative of a Composite Function

Suppose \( f(1) = 4 \) and \( f'(1) = 3 \). If

\[ g(x) = \sqrt{f(x)} \]

then \( g'(1) \) equals

  • A \( \frac{3}{4} \)
  • B. \( \frac{3}{2} \)
  • C. \( \frac{1}{2\sqrt{3}} \)
  • D. \( \frac{2}{3} \)
  • E. \( \frac{1}{4} \)

Solution Steps

\( g(x) = [f(x)]^{1/2} \)

Treat it like \( u^n \)

deriv: \( n u^{n-1} \cdot \frac{du}{dx} \)

\[ g'(x) = \frac{1}{2} [f(x)]^{-1/2} \cdot f'(x) \]\[ g'(x) = \frac{1}{2} \frac{1}{\sqrt{f(x)}} \cdot f'(x) \]\[ g'(1) = \frac{1}{2} \frac{1}{\sqrt{f(1)}} \cdot f'(1) = \frac{1}{2} \frac{1}{\sqrt{4}} \cdot 3 = \frac{3}{4} \]
PAGE 8

Calculus Problem: Implicit Differentiation

The slope of the line tangent to the curve described by the implicit function \( y^3x + y^2x^2 = 6 \) at \( (2, 1) \) is

  • A. \( -\frac{3}{2} \)
  • B. \( -1 \)
  • C. \( -\frac{3}{14} \)
  • D. 0
  • E \( -\frac{5}{14} \)

Solution Steps

find \( \frac{dy}{dx} \) at \( x=2, y=1 \)

differentiate \( y^3x + y^2x^2 = 6 \) implicitly

\[ \frac{d}{dx}(y^3x + y^2x^2) = \frac{d}{dx}(6) \]\[ \frac{d}{dx}(y^3x) + \frac{d}{dx}(y^2x^2) = \frac{d}{dx}(6) \]

[ product rule ]

\[ \left[ (y^3)(1) + (x)(3y^2)\frac{dy}{dx} \right] + \left[ (y^2)(2x) + (x^2)(2y)\frac{dy}{dx} \right] = 0 \]

plug in \( x=2, y=1 \)

\[ (1 + 6\frac{dy}{dx}) + (4 + 8\frac{dy}{dx}) = 0 \]\[ 14\frac{dy}{dx} = -5 \]
\[ \frac{dy}{dx} = -\frac{5}{14} \]
PAGE 9

F21 Exam 2

Problem #2

\[ y = \frac{t^2 e^{-t}}{t^{1/2} + 1} \]

Find \( \frac{dy}{dt} \) at \( t = 1 \).

Quotient rule to start:

\[ y' = \frac{(t^{1/2} + 1) \frac{d}{dt}(t^2 e^{-t}) - (t^2 e^{-t})(\frac{1}{2} t^{-1/2})}{(t^{1/2} + 1)^2} \]

Product Rule Annotation:

\[ \frac{d}{dt}(t^2 e^{-t}) = t^2 \cdot e^{-t} \cdot (-1) + e^{-t} \cdot 2t \]
\[ = \frac{(t^{1/2} + 1)(-t^2 e^{-t} + 2t e^{-t}) - (t^2 e^{-t})(\frac{1}{2} t^{-1/2})}{(t^{1/2} + 1)^2} \]

Plug in \( t = 1 \):

\[ = \frac{(2)(-e^{-1} + 2e^{-1}) - (e^{-1})(\frac{1}{2})}{4} \]\[ = \frac{2e^{-1} - \frac{1}{2}e^{-1}}{4} = \frac{\frac{3}{2}e^{-1}}{4} = \frac{3}{8e} \]
\[ \frac{3}{8e} \]
PAGE 10

F21 Exam 2

Problem #8

\[ f(x) = \cos(\pi e^{3x}) \]

Find \( f'(\frac{1}{3} \ln \frac{1}{2}) \).

Chain Rule Reminders:

  • \( \frac{d}{dx} \cos(u) = -\sin(u) \frac{du}{dx} \)
  • \( \frac{d}{dx} e^u = e^u \frac{du}{dx} \)
\[ f'(x) = -\sin(\pi e^{3x}) \cdot \frac{d}{dx}(\pi e^{3x}) \]\[ = -\sin(\pi e^{3x}) \cdot \pi e^{3x} \cdot 3 \]

Evaluate at \( x = \frac{1}{3} \ln \frac{1}{2} \):

Note: \( e^{3 \cdot \frac{1}{3} \ln \frac{1}{2}} = e^{\ln \frac{1}{2}} = \frac{1}{2} \)

\[ f'(\frac{1}{3} \ln \frac{1}{2}) = -\sin(\pi (e^{3 \cdot \frac{1}{3} \ln \frac{1}{2}})) \cdot \pi e^{3 \cdot \frac{1}{3} \ln \frac{1}{2}} \cdot 3 \]\[ = -\sin(\frac{\pi}{2}) \cdot 3\pi \cdot \frac{1}{2} \]

Since \( \sin(\frac{\pi}{2}) = 1 \):

\[ = -1 \cdot \frac{3\pi}{2} = -\frac{3\pi}{2} \]