
# 1.4E Exercises

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### Limit at infinity

Exercise $$\PageIndex{1}$$

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

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

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

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

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

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

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

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

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

$$\displaystyle = 0$$

Exercise $$\PageIndex{10}$$

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

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

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

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

### Exercise $$\PageIndex{12}$$: Horizontal Asymptotes

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

$$y=3$$ and $$y=-3$$.