Is $ \lim_{x\to \infty}\frac{x^2}{x+1} $ equal to $\infty$ or to $1$? What is 
$ \lim_{x\to \infty}\frac{x^2}{x+1}$? 
When I look at the function's graph, it shows that it goes to $\infty$, but If I solve it by hand it shows that the $\lim \rightarrow 1$
$ \lim_{x\to \infty}\frac{x^2}{x+1} =$ $ \lim_{x\to \infty}\frac{\frac{x^2}{x^2}}{\frac{x}{x^2}+\frac{1}{x^2}} =$ $\lim_{x\to \infty}\frac{1}{\frac{1}{x}+\frac{1}{x^2}}=$$ \lim_{x\to \infty}\frac{1}{0+0} = \ 1$  
what is wrong here?
 A: The answer is $+\infty$. Note that $\lim_ {x\to\infty}\frac1{\frac1x+\frac1{x^2}}=+\infty$, not $1$.
A: What is wrong is 


*

*to write "$\lim_{x\to \infty} \frac{1}{0+0}$": an expression such as $\frac{1}{0+0}$ doesn't make sense.

*to write that "$\lim_{x\to \infty} \frac{1}{0+0}$" is equal to $1$. If there is a way to give a meaning to this limit, it will certainly not be $1$ but rather $+\infty$ or $-\infty$ depending on the sign of the denominator.

*to constantly mix up the two kinds of sentences "the function tends to infinity" and "the limit of the function is equal to infinity".
A: $$\lim_{x \rightarrow \infty} \frac{x^2}{x+1}=\lim_{x \rightarrow \infty} x\cdot \frac{1}{1+\frac1x}=\infty$$
A: You  cannot always reinvent the wheel! It's a  basic result that the limit at infinity of a rational function is the  limit of the ratio of the leading terms of its numerator and denominator. 
In other terms, here:
$$\lim_{x\to\infty}\frac{x^2}{x+1}=\lim_{x\to\infty}\frac{x^2}{x}=\lim_{x\to\infty}x.$$
A: Writing the fraction as follows may clear it up:
$$
\frac{x^2}{x+1}
= \frac{x^2-1+1}{x+1}
= \frac{x^2-1}{x+1} + \frac{1}{x+1}
= x-1 + \frac{1}{x+1}
\to \infty
$$
A: Hint: Write $$x\cdot \frac{1}{1+\frac{1}{x}}$$
