Is $ \lim_{x\to \infty}\frac{x^2}{x+1} $ equal to $\infty$ or to $1$?

Question

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?

Answer 1

The answer is $+\infty$. Note that $\lim_ {x\to\infty}\frac1{\frac1x+\frac1{x^2}}=+\infty$, not $1$.

Answer 2

Hint: Write $$x\cdot \frac{1}{1+\frac{1}{x}}$$

Answer 3

$$\lim_{x \rightarrow \infty} \frac{x^2}{x+1}=\lim_{x \rightarrow \infty} x\cdot \frac{1}{1+\frac1x}=\infty$$

Answer 4

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: $$\lim_{x\to\infty}\frac{x^2}{x+1}=\lim_{x\to\infty}\frac{x^2}{x}=\lim_{x\to\infty}x.$$

Answer 5

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".

Answer 6

Writing the fraction as follows may clear it up: $$ \frac{x^2}{x+1} = \frac{x^2-1+1}{x+1} = \frac{x^2-1}{x+1} + \frac{1}{x+1} = x-1 + \frac{1}{x+1} \to \infty $$