1. ## infinite limits

I'm still trying to grasp the idea of limits as a whole, and infinite limits, individually. I really don't know where to start when it comes to multiplying the numerator and denom by a certain power of 1/x to talk myself through the problem... Here are a few of my homework problems, any addition info to show me how you work through the problem would be greatly appreciated.

1. lim as x approaches infinity of x + x^3 + x^5/1 - x^2 + x^4 =

2. lim as t approaches negative infinity of x^2 + 2/t^3 + t^2 -1

3. lim as x approaches infinity of x + 2/ sq rt (9x^2 + 1)

4. lim as x approaches infinity of [(sq rt(9x^2 + x)) - 3x]
*(9x^2 + x) is the only thing supposed to be under the radical.

2. Hello, obsmith08!

The rule I've taught my students is:
Divide numerator and denominator by the highest power of the variable in the denominator.

$1)\;\;\lim_{x\to\infty}\frac{x + x^3 + x^5}{1 - x^2 + x^4}$

Divide top and bottom by $x^4\!:\;\;\frac{\dfrac{x}{x^4} + \dfrac{x^3}{x^4} + \dfrac{x^5}{x^4}}{\dfrac{1}{x^4} - \dfrac{x^2}{x^4} + \dfrac{x^4}{x^4}}$ . $= \;\frac{\dfrac{1}{x^3} + \dfrac{1}{x} + x}{\dfrac{1}{x^4} - \dfrac{1}{x^2} + 1}$

. . Then: . $\lim_{x\to\infty} \frac{\frac{1}{x^3} + \frac{1}{x} + x}{\frac{1}{x^4} - \frac{1}{x^2} + 1} \;=\;\frac{0+0+\infty}{0-0+1} \;=\;\infty$

$2)\;\;\lim_{x\to-\infty}\frac{x^2 + 2}{x^3 + x^2 -1}$

Divide top and bottom by $x^3\!:\;\;\frac{\dfrac{x^2}{x^3} + \dfrac{2}{x^3}}{\dfrac{x^3}{x^3} + \dfrac{x}{x^3} - \dfrac{1}{x^3}} \;=\;
\frac{\dfrac{1}{x} + \dfrac{2}{x^3}}{1 + \dfrac{1}{x} - \dfrac{1}{x^3}}$

.Then: . $\lim_{x\to-\infty}\frac{\dfrac{1}{x} + \dfrac{2}{x^3}}{1 + \dfrac{1}{x} - \dfrac{1}{x^3}}\;=\;\frac{0+0}{1+0-0} \;=\;0$

$3)\;\;\lim_{x\to\infty}\frac{x + 2}{\sqrt{9x^2 + 1}}$

$\text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{x+2}{x}}{\dfrac{\sqrt{9x^2+1 }}{x}} \;=\;\frac{\dfrac{x}{x} + \dfrac{2}{x}} {\dfrac{\sqrt{9x^2+1}}{\sqrt{x^2}}} \;=\;\frac{1+\dfrac{2}{x}} {\sqrt{\dfrac{9x^2}{x^2} + \dfrac{1}{x^2}}}$ . $=\;\frac{1 + \dfrac{2}{x}} {\sqrt{9 + \dfrac{1}{x^2}}}$
. . Then: . $\lim_{x\to\infty}\frac{1 + \dfrac{2}{x}} {\sqrt{9 + \dfrac{1}{x^2}}} \;=\;\frac{1+0}{\sqrt{9+0}} \;=\;\frac{1}{3}$

$4)\;\;\lim_{x\to\infty}\left[\sqrt{9x^2 + x} - 3x\right]$
This one requires a different technique.

Multiply top and bottom by the conjugate:

. . $\frac{\sqrt{9x^2+x} - 3x}{1}\cdot\frac{\sqrt{9x^2+x} + 3x}{\sqrt{9x^2+x} + 3x} \;=\;\frac{(9x^2+x) - 9x^2}{\sqrt{9x^2+x} + 3x} \;=\;\frac{x}{\sqrt{9x^2+x}+3x}$

$\text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{x}{x}} {\dfrac{\sqrt{9x^2+x}}{\sqrt{x^2}} + \dfrac{3x}{x}} \;=\;\frac{1} {\sqrt{\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}} + 3} \;=\;\frac{1}{\sqrt{9+\dfrac{1}{x}} + 3}$

. . Then: . $\lim_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}} + 3} \;=\;\frac{1}{\sqrt{9+0} + 3} \;=\;\frac{1}{3+3} \;=\;\frac{1}{6}$