# Thread: How do you evaluate limits at infinity?

1. ## How do you evaluate limits at infinity?

I have a couple of questions asking me to evaluate limits at infinity, but I am a bit confused.

These are the questions given in the example:

Question: Evaluate limx->infinity (1) / (x+1)

Question: Evaluate limx->infinity (x^3 + x^2 - 3) / (4x^3)

So I'm not sure what the process is to solve questions like these so that I can use the same process in my homework questions.

Any help is appreciated.

2. Usually, one has to know of a few properties of division by infinity, infinity as an exponent etc. You can easily find a comprehensive list of such properties either online or in any calculus textbook. For your particular cases, "force factorization" will work.

a) limx ->infinity 1/(x+1)=x(1/x)/x(1+1/x)=0 Here I factored both the denominator and the numerator by x; 1/x as x approaches inifnity is 0, so 0/(1+0)=0
b)limx ->inifnity (x^3+x^2-3)/(4x^3)=x^3(1+(1/x)-(3/x^2))/x^3(4)=(1+0-0)/4=1/4

3. Originally Posted by iluvmathbutitshard
I have a couple of questions asking me to evaluate limits at infinity, but I am a bit confused.

These are the questions given in the example:

Question: Evaluate limx->infinity (1) / (x+1)

Question: Evaluate limx->infinity (x^3 + x^2 - 3) / (4x^3)

So I'm not sure what the process is to solve questions like these so that I can use the same process in my homework questions.

Any help is appreciated.
$\displaystyle \displaystyle\lim_{x\rightarrow\infty}\frac{1}{x+1 }=\lim_{x\rightarrow\infty}\frac{1}{x\left(1+\frac {1}{x}\right)}=\lim_{x\rightarrow\infty}\frac{1}{x }\left(\frac{1}{1+\frac{1}{x}}\right)$

Now use the fact that $\displaystyle \displaystyle\lim_{x\rightarrow\infty}\frac{1}{x}= 0$

$\displaystyle \displaystyle\lim_{x\rightarrow\infty}\frac{x^3+x^ 2-3}{4x^3}=\lim_{x\rightarrow\infty}\left(\frac{x^3} {4x^3}+\frac{x^2}{4x^3}-\frac{3}{4x^3}\right)=\lim_{x\rightarrow\infty}\le ft(\frac{1}{4}+\frac{1}{4x}-\frac{3}{4x^3}\right)$

4. That's more clear than text math, thanks Archie.