# Math Help - calculate limit

1. ## calculate limit

2. There are a couple of different ways to show that the limit is $0$.

$\lim_{x\to\infty}\frac{\ln(x^2+x+1)}{x}=\lim_{x\to \infty}\frac{\ln\left(1+\frac{1}{x}+\frac{1}{x^2}\ right)+\ln x^2}{x}=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1} {x}+\frac{1}{x^2}\right)+2\ln x}{x}$

You should know that $\lim_{x\to \infty}\frac{\ln x}{x}=0$

Or you can use L'Hospitals rule twice with $f(x)=\ln(x^2+x+1)$ and $g(x)=x$

$\lim_{x\to \infty}\frac{f(x)}{g(x)}=\lim_{x\to \infty}\frac{f'(x)}{g'(x)}=\lim_{x\to \infty}\frac{f''(x)}{g''(x)}=\lim_{x\to \infty}\frac{2}{3x^2+2x+1}$

3. Originally Posted by spec
there are a couple of different ways to show that the limit is $0$.

$\lim_{x\to\infty}\frac{\ln(x^2+x+1)}{x}=\lim_{x\to \infty}\frac{\ln\left(1+\frac{1}{x}+\frac{1}{x^2}\ right)+\ln x^2}{x}=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1} {x}+\frac{1}{x^2}\right)+2\ln x}{x}$

you should know that $\lim_{x\to \infty}\frac{\ln x}{x}=0$

or you can use l'hospitals rule twice with $f(x)=\ln(x^2+x+1)$ and $g(x)=x$

$\lim_{x\to \infty}\frac{f(x)}{g(x)}=\lim_{x\to \infty}\frac{f'(x)}{g'(x)}=\lim_{x\to \infty}\frac{f''(x)}{g''(x)}=\lim_{x\to \infty}\frac{2}{3x^2+2x+1}$
thank you
if x to -infty ?

4. Same result.