Limit approaching infinity-related question

Question

Why is $$\lim_{x\to\infty}\frac{x^2}{1+x^2}=1?$$

Answer 1

We can explain it in two different manners.

First way: Divide everything by $x^2$ and simply get $\lim\limits_{x\to\infty}\frac {1}{1+\frac 1 {x^2}}=1$

Second way: $\frac {x^2}{x^2+1}=\frac {x^2+1}{x^2+1}-\frac{1} {x^2+1}=1-\frac 1 {x^2+1}$ and since $\lim\limits_{x\to\infty}\frac 1 {x^2+1}=0$,$\lim\limits_{x\to\infty}\frac {1}{1+\frac 1 {x^2}}=1$

Both ways lead to the result that $$\lim\limits_{x\to\infty}\frac {x^2}{x^2+1}=1 $$

Answer 2

Note that:

$$\frac{x^2}{1+x^2}=\frac{x^2\cdot\frac1{x^2}}{(1+x^2)\cdot\frac1{x^2}}=\frac{1}{\frac1{x^2}+1}$$

Now let $x\to\infty$ and we have as wanted.