1.4E Exercises

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Page ID: 10660

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Limit at infinity

Exercise $\PageIndex{1}$

Find $ \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x$

Answer

To find $ \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x$, First multiply by the conjugate

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

Then reduce the numerator

$ \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3}{\sqrt{x^2+3}+x}$

Now divide by $ \sqrt{x^2}=|x|$,

$ \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3/|x|}{(\sqrt{x^2+3}+x)/ \sqrt{x^2}}$

Since $x \to \infty$, replace $|x|$ by $x$.

Thus, \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3/x}{(\sqrt{1+3/\sqrt{x^2}}+x/x)

This result will be:

$ \displaystyle = 3(0)/ 2$

$ \displaystyle = 0$

Exercise $\PageIndex{2}$

Find $ \displaystyle \lim_{x \to - \infty} \frac{\sqrt{x^4+6x}}{x^2-6}$

Answer

Factor out the highest degree variable in the numerator and the denominator, $ \displaystyle \sqrt{x^4}=|x^2|=x^2$.

$ \displaystyle \lim_{x \to - \infty} \frac{\sqrt{x^4+6x}}{x^2-6} = \lim_\limits{x \to - \infty} \frac{x^2\sqrt{1 + \frac{6x}{x^3}}}{(x^2)(1- \frac{6}{x^2})}$

Evaluate:

$ \displaystyle = \frac{\sqrt{1+0}}{1-0}$

$ \displaystyle = \frac{1}{1}$

$ \displaystyle = 1$

Exercise $\PageIndex{3}$

Find $ \displaystyle \lim\limits_{t \to \infty} \frac{7-t}{\sqrt{2+2t^2}}$

Answer

Factor out the highest degree variable in the numerator and denominator

= $ \displaystyle \lim\limits_{t \to\infty} \frac{t (\frac{7}{t} -1)}{|t|\sqrt{\frac{2}{t^2}+2}}$

Evaluate:

$ \displaystyle = \frac{0-1}{\sqrt{0+2}}$

$ \displaystyle = \frac{-1}{\sqrt{2}}$

Exercise $\PageIndex{4}$

Find $ \displaystyle \lim_{t \to - \infty} \frac{7-t}{\sqrt{2+2t^2}}$

Answer

Factor out the highest degree variable in the numerator and denominator

$ \displaystyle =\lim_{t \to \infty} \frac{(t)(\frac{7}{t}-1)}{|t|\sqrt{\frac{2}{t^2}+2}}$

$ \displaystyle = \frac{0-1}{(-1)\sqrt{0+2}}$

$ \displaystyle = \frac{1}{\sqrt{2}}$

Exercise $\PageIndex{5}$

Find $ \displaystyle \lim_{y \to - \infty} \frac{\sqrt{4y^2-2}}{2-y}$

Answer

Factor out the highest degree variable in the numerator and denominator

$ \displaystyle = \lim_{y \to -\infty} \frac{y\sqrt{4-\frac{2}{y^2}}}{|y|(\frac{2}{y}-1)}$

When $y \to -\infty $ $ |y | $approaches to $-\infty$.

Evaluate:

$ \displaystyle \frac{\sqrt{4-0}}{-(0-1)}$

$ \displaystyle \frac{2}{1}$

$ \displaystyle 2$

Exercise $\PageIndex{6}$

Find $ \displaystyle \lim_{s \to \infty} \sqrt[5]{ \frac{5s^8 - 4s^3}{2s^8-1}}$

Answer

Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

= $ \displaystyle \lim_{s \to \infty} \sqrt[5]{\frac{5s^8}{2s^8}}$

$= \displaystyle \sqrt[5]{\frac{5}{2}}$

Exercise $\PageIndex{7}$

Find $ \displaystyle \lim_{x \to \infty} \sqrt[3]{ \frac{3+2x+5x^2}{1+4x^2}}$

Answer

Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

$ \displaystyle = \lim_{x \to \infty} \sqrt[3]{\frac{5x^2}{4x^2}}$

$ \displaystyle = \sqrt[3]{\frac{5}{4}}$.

Exercise $\PageIndex{8}$

Find $ \displaystyle \lim_{t \to \infty} \frac {11-t^5}{11t^5 +1}$

Answer

Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

$ \displaystyle = \lim_{t \to \infty} \frac {-t^5}{11t^5}$

$ \displaystyle = \frac {-1}{11}$

Exercise $\PageIndex{9}$

Find $ \displaystyle \lim_{x \to -\infty} \frac{3}{x-3}$

Answer: $ \displaystyle = 0$

Exercise $\PageIndex{10}$

Find $ \displaystyle \lim_{x \to \infty} \frac{3x^2-2x}{5x^2+1}$

Answer

Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

$ \displaystyle = \lim_{x \to -\infty} \frac{3x^2}{5x^2}$

$ \displaystyle = \frac{3}{5}$.

Exercise $\PageIndex{11}$

Find $\displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2+6y-8}}{5-y}$.

Answer

$\displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2+6y-8}}{5-y}= \displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2}}{-y}$

$\displaystyle\lim_{y \to -\infty} \frac{\frac{\sqrt{7y^2}}{|y|}}{\frac{-y}{|y|}}= \displaystyle\lim_{y \to -\infty} \frac{\frac{\sqrt{7y^2}}{\sqrt{y^2}}}{\frac{-y}{-y}} = \sqrt{7} $

Exercise $\PageIndex{12}$: Horizontal Asymptotes

Find horizontal asymptote(s) of the function $$ y= \displaystyle \frac{5+3x}{ \sqrt{x^2+x-4}}.$$

Answer: $y=3$ and $y=-3$.

Contributors

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)

Limit at infinity

Exercise \(\PageIndex{12}\): Horizontal Asymptotes

Contributors