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1) $ \lim x^4/(1+x^2)=$ infinite when $x$ goes to infinity

I know that I need to prove that $x^4/(1+x^2)>$ than something divergent.

2) $\lim x^2/(1+x^2)=1$ when $x$ goes to infinity. Is it enough to say let $x_n$ be a sequence that goes to infinity, $x^2/(1+x^2)=1/(1/x^2+1)$ then $f(x_n)=1$ ??

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For the first question it's only required that the inequality is eventually true. For example $x^4/(1+x^2) > x^2/2$ as soon as $1+x^2 > 2$ (ie $x>1$).

For the second question I'm not sure what you're aiming at here. If $\lim_{x\to\infty} f(x)$ exists then certainly also $\lim_{n\to\infty}f(x_n)$ also exists with the same value (as $x_n\to\infty$). However the reverse is not true, consider for example $f(x) = \sin x$ and $x_n = n\pi$ then we have $f(x_n)=0$ so $\lim_{n\to\infty} f(x_n) = 0$, but $f(x)$ itself has no limit as $x\to\infty$.

If you want to prove that $x^2/(1+x^2)\to1$ you could try to use a similar technique as in the first, but it's easier to rewrite the expression (your rewriting $1/(1+1/x^2)$ works to):

$${x^2\over1+x^2} = {x^2+1-1\over x^2 +1 } = {x^2+1\over x^2+1}-{1\over x^2+1} = 1 - {1\over 1+x^2}$$

You could perhaps use the first technique to realize that $1/(1+x^2)\to0$.

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    Thank you so much that what I need :-) – 2017-02-20
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    I know I am late , but this question was in the exam yesterday if $ 1+x^2 >2 then x^4/(1+x^2) < x^2 /2 $ and this does not help we need the right greater than the left ,so I suppose that $1+x^2 < x^2+x^2 then x^4/(1+x^2) > x^3/2x^2 =x/2$ which goes to infinite I.e diverges – 2017-03-13