Limits with infinity exercises

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Here is an opportunity for you to practice evaluating limits that involve infinity.

Your goal in this exercise is to evaluate $\displaystyle \lim _{x \to -3} \frac {1}{(x+3)^3}$ (if it exists). First, consider the corresponding one-sided limits:

$\lim _{x \to -3^-} \frac {1}{(x+3)^3} \text { and } \lim _{x \to -3^+} \frac {1}{(x+3)^3}.$

As $x$ approaches $-3$ from the left:

The numerator approaches $\answer {1}$ , which is a positivenegative number.
$(x+3)^3$ is and approaches $\answer {0}.$

Therefore,

$\lim _{x \to -3^-} \frac {1}{(x+3)^3} = \answer {-\infty }.$

As $x$ approaches $-3$ from the right:

The numerator approaches $\answer {1}$ , which is a positivenegative number.
$(x+3)^3$ is and approaches $\answer {0}.$

Therefore,

$\lim _{x \to -3^+} \frac {1}{(x+3)^3} = \answer {\infty }.$

Putting all of this information together, you can conclude that

$\lim _{x \to -3} \frac {1}{(x+3)^3} = \answer {DNE}.$

From the above limit calculations, you can say that $\frac {1}{(x+3)^3}$

doesdoes not have a horizontalvertical asymptote at $x =-3$ .

Your goal in this exercise is to evaluate $\displaystyle \lim _{x \to 7} \frac {-3x+2}{(x-7)^6}$ (if it exists). First, consider the corresponding one-sided limits:

$\displaystyle \lim _{x \to 7^-} \frac {-3x+2}{(x-7)^6} \text { and } \displaystyle \lim _{x \to 7^+} \frac {-3x+2}{(x-7)^6}.$

As $x$ approaches $7$ from the left:

The function $-3x+2$ approaches $\answer {-19}$ , which is a positivenegative number.
$(x-7)^6$ is and approaches $\answer {0}$ .

Therefore,

$\displaystyle \lim _{x \to 7^-} \frac {-3x+2}{(x-7)^6} = \answer {-\infty }.$

As $x$ approaches $7$ from the right:

The function $-3x+2$ approaches $\answer {-19}$ , which is a positivenegative number.
$(x-7)^6$ is and approaches $\answer {0}.$

Therefore,

$\displaystyle \lim _{x \to 7^+} \frac {-3x+2}{(x-7)^6} = \answer {-\infty }.$

Putting all of this information together, you can conclude that

$\displaystyle \lim _{x \to 7} \frac {-3x+2}{(x-7)^6} = \answer {-\infty }.$

From the above limit calculations, you can say that $\frac {-3x+2}{(x-7)^3}$

doesdoes not have a horizontalvertical asymptote at $x =7$ .

Consider the function $f(x) = \frac {x^2+x-6}{x^2-3x+2}$ . The two $x$ -values at which $f(x)$ may have a vertical asymptote are

On this problem, it will help to factor the numerator and denominator of $f(x)$ .

$x = -2, 6$ $x = 1, 2$ $x = -3, 1$ $x = 2, -3$

Good thinking! For a rational function like $f(x)$ , there could be an asymptote at all the $x$ values where $f$ is undefined. The only way for $f$ to be undefined is if the denominator of $f$ is 0. Since $x^2 -3x + 2 = (x-1)(x-2)$ , the $x$ values that make the denominator 0 are $x = 1, 2$ , so these are the potential values at which $f$ may have a vertical asymptote.

Now, determine if $x = 1$ and $x = 2$ are asymptotes of $f$ or not.

As $x$ approaches $1$ from the left:

The function $x^2 + x -6$ approaches $\answer {-4}$ , which is a positivenegative number.
$x^2 - 3x + 2$ is and approaches $\answer {0}$ .

Therefore,

$\displaystyle \lim _{x \to 1^-} \frac {x^2+x-6}{x^2-3x+2} = \answer {-\infty }$

and based on this limit information, you can conclude that

$x = 1$ is a vertical asymptote of $f$ . $x = 1$ is not a vertical asymptote of $f$ . $x = 1$ may or may not be a vertical asymptete of $f$ . There isn’t enough information to decide yet.

Alright, now you know that $x=1$ is a vertical asymptote of $f$ . What about $x=2$ ? Well,

$\displaystyle \lim _{x \to 2^-} \frac {x^2+x-6}{x^2-3x+2} = \answer {5}$

and based on this limit information, you can conclude that

$x = 2$ is a vertical asymptote of $f$ . $x = 2$ is not a vertical asymptote of $f$ . $x = 2$ may or may not be a vertical asymptote of $f$ . There isn’t enough information to decide yet.

Because the limit information we have so far isn’t enough to make any conclusions, evaluate the other one-sided limit:

$\displaystyle \lim _{x \to 2^+} \frac {x^2+x-6}{x^2-3x+2} = \answer {5}.$

Now, from all information you have obtained about the limit of $f(x)$ at $x=2$ , you can conclude that

$x = 2$ is a vertical asymptote of $f$ . $x = 2$ is not a vertical asymptote of $f$ . $x = 2$ may or may not be a vertical asymptote of $f$ . There isn’t enough information to decide yet.

It turns out that $\displaystyle \lim _{x \to 2} \frac {x^2+x-6}{x^2-3x+2} = 5$ while $f(2)$ does not exist. This indicates that there is a hole at $x=2$ and not a vertical asymptote! In fact, if you factor $f(x)$ , you’ll realize that a factor of $(x-2)$ appears precisely once in the numerator and denominator. Because of this, when you evaluate $\displaystyle \lim _{x \to 2} \frac {x^2+x-6}{x^2-3x+2}$ , the $(x-2)$ ’s will cancel out, and you’ll find that the limit exists. This kind of ’cancellation set-up’ is a tell-tale sign that the function at hand has a hole at $x=2$ rather than a vertical asymptote.

Evaluate the limit.

$\displaystyle \lim _{x \to \infty } \frac {3x +7 - x^5}{x^4 + 8x^2 -3} = \answer {-\infty }$

To evaluate the above limit, you could multiply the numerator and denominator of $\frac {3x +7 - x^5}{x^4 + 8x^2 -3}$ by $\cfrac {1}{x^{\answer {4}}}.$ (Type in the smallest exponent that would work)

Evaluate the limit.

$\displaystyle \lim _{x \to -\infty } \frac {5x^6 + 3.5x^3 -70x^2 + 2\pi }{3x - \frac {1}{3}x^3+ 6x^6} = \answer {\frac {5}{6}}$

To evaluate the above limit, you could multiply the numerator and denominator of $\frac {5x^6 + 3.5x^3 -70x^2 + 2\pi }{3x - \frac {1}{3}x^3+ 6x^6}$ by $\cfrac {1}{x^{\answer {6}}}.$ (Type in the smallest exponent that would work)

Evaluate the limits.

$\displaystyle \lim _{x \to \infty } \frac {x - 2x^2 + 3x^3 -4x^4}{7x+5-x^5} = \answer {0}$

$\displaystyle \lim _{x \to -\infty } \frac {x - 2x^2 + 3x^3 -4x^4}{7x+5-x^5} = \answer {0}$

To evaluate the above limits, you could multiply the numerator and denominator of $\frac {x - 2x^2 + 3x^3 -4x^4}{7x+5-x^5}$ by $\cfrac {1}{x^{\answer {5}}}.$ (Type in the smallest exponent that would work)

Evaluate the limit.

$\displaystyle \lim _{x \to \infty } \frac {\sqrt {2x^2+5}}{-3x} = \answer {-\frac {\sqrt {2}}{3}}$

Based on this limit, you can conclude that $\frac {\sqrt {2x^2+5}}{-3x}$ has a horizontal asymptote of $\answer {y} \ = \answer {-\frac {\sqrt {2}}{3}}$ .

The function $f(x) = \frac {-1 + 2|x|}{x}$ has a vertical asymptote of $\answer {x} \ = \answer {0}$ . This function also has two horizontal asymptotes: One below the $x$ -axis, $\answer {y} \ = \answer {-2}$ , and one above the $x$ -axis, $\answer {y} \ = \answer {2}$ .

Use the limit definition of vertical and horizontal asymptote in order to answer this question. (Yes, that means you’ll have to calculate some limits!)

A function can have a maximum of $\answer {2}$ horizontal asymptotes.

Could you explain to a fellow classmate why a function can have no more than 2 vertical asymptotes using the limit definition of a horizontal asymptote?

Consider the statement below, and then indicate whether it is sometimes, always, or never true.

‘‘A function, $f(x)$ , can have both a vertical and a horizontal asymptote of $x=-3$ .”

This statement is

sometimesalwaysnever true.