6.2E Exercises

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Finding Limits From Graphs

By looking at the graph of $f(x)$ below, find the indicated limits.

$limitsgraph.png$

\[ \lim_{x \rightarrow -5} f(x), \quad \lim_{x \rightarrow 0^-} f(x), \quad \lim_{x \rightarrow 0^+} f(x), \quad \lim_{x \rightarrow 0} f(x), \notag \]

\[ \lim_{x \rightarrow 4^-} f(x), \quad \lim_{x \rightarrow 4^+} f(x), \quad \lim_{x \rightarrow 4} f(x), \quad \lim_{x \rightarrow -\infty} f(x) \notag\]

Answer

$ \lim_{x \rightarrow -5} f(x) = 4 $
$ \lim_{x \rightarrow 0^-} f(x) = 1 $
$ \lim_{x \rightarrow 0^+} f(x) = -2 $
$\lim_{x \rightarrow 0} f(x) $ DNE.
$ \lim_{x \rightarrow 4^-} f(x) = 4 $
$\lim_{x \rightarrow 4^+} f(x) = 4 $
$ \lim_{x \rightarrow 4} f(x) = 4 $
$ \lim_{x \rightarrow -\infty} f(x) = \infty $

Identifying Asymptotes From Graphs

By looking at the graphs of rational functions below, identify any vertical, horizontal, or slant asymptotes.

1.	2.
$aa.png$	$b.png$
3.	4.
$c.png$	$d.png$

Answer

V.A. $x = 0$, Slant $y = 2x$.
V.A. $ x = 1$, H.A. $y = -2$.
V.A.s $ x = -1, x = 2$, H.A. $y = 3 $.
V.A. $ x = 1$, Slant $ y = x+1 $.

Algebraically Finding Vertical Asymptotes

Find any vertical asymptotes of the rational functions by analyzing appropriate limits.

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

2. $ f(x) = \dfrac{ x}{x+2} $

3. $ f(x) = \dfrac{-2x^2+4x-1}{(x-1)^2} $

4. $ f(x) = \dfrac{ (x+2)(x+3)(x-3)}{(x+2)(x-2)(x+1)} $

5. $ f(x) = \dfrac{ 2x^5+x^2}{x^4} $

Answer

HINT: If you can't get started, the steps are 1) identify domain issues and 2) analyze the limits as $x$ approaches those locations to see if any of them have a limit of $\pm \infty$. Go do them and come back to check your answers.

$ x = 2$ and $x = 3$ are V.A.s.
$ x = -2 $ is a V.A.
$ x = 1$ is a V.A. (As $x \rightarrow 1$, the numerator heads toward 1 while the denominator heads toward 0, making the fraction blow up.)
$ x = -1 $ and $ x= 2$ are V.A.s. (When you consider $x \rightarrow -2$, notice that by cancelling in the top and bottom as possible, the fraction's value heads toward $-\frac{5}{4} $, not $\pm \infty$.)
$ x = 0 $ is a V.A. (Write $ \dfrac{ 2x^5 + x^2}{x^4} $ as $ 2x + \frac{1}{x^2} $. As $x \rightarrow 0$, the first term goes away but the second term blows up forever.)

Algebraically Finding Horizontal Asymptotes

Find any horizontal asymptotes of the rational functions by analyzing appropriate limits.

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

2. $ f(x) = \dfrac{ x}{x+2} $

3. $ f(x) = \dfrac{-2x^2+4x-1}{(x-1)^2} $

4. $ f(x) = \dfrac{ (x+2)(x+3)(x-3)}{(x+2)(x-2)(x+1)} $

5. $ f(x) = \dfrac{ 2x^5+x^2}{x^4} $

Answer

HINT: If you can't get started, the limits you need to analyze are as $x \rightarrow \infty$ and/or $x \rightarrow -\infty$. Remember the trick of dividing top and bottom by the highest power of $x$ from the denominator.

H.A. $ y = 0$.
H.A. $ y = 1$.
H.A. $ y = -2$.
H.A. $ y = 1$.
No H.A.

Algebraically Finding Slant Asymptotes

Find any slant asymptotes of the rational functions by analyzing appropriate limits.

1. $ f(x) = \dfrac{x^2+1}{x} $

2. $ f(x) = \dfrac{ 2x^5+x^2}{x^4} $

2. $ f(x) = \dfrac{ 3x^3+2x^2-x+1}{x^2+1} $

Answer

HINT: The question is, does the function behave like a linear function as $x \rightarrow \pm \infty$? Use polynomial long division (or splitting the fraction and cancelling) to see.

$ y = x $
$ y = 2x $
$ y = 3x + 2$

Limit Application

The zombie apocalypse has begun (womp womp). Currently, the spread of the outbreak is modeled by the function $P(d) = \frac{1}{2}d^3 - d + 1000 $, giving the number of zombies on day $d$ after the outbreak started. If nothing changes and time goes on with this growth function unchecked, will the number of zombies (a) decrease and go extinct, (b) level out at a certain number, or (c) grow until they take over the Earth?

Answer

If time is going on forever, we're looking at sending $ d\rightarrow \infty$. As $d$ grows, the term $\frac{1}{2}d^3$ will grow very quickly. Even though the $ -d$ term is subtracting off some zombies, it can't offset the quick growth of the cubic term! Think about it...

\[ \frac{1}{2}2^3 - 2 = 2, \quad ... \quad \frac{1}{2}10^3 - 10 = 490, \quad ... \quad \frac{1}{2} 100^3 - 100 = 499,900, \quad ... \notag \]

Believe me? So as time goes on, the zombies take over the Earth. (rip)

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