# Proving a limit $\lim\limits_{x\to \infty}\frac{x-3}{x^2 +1}$ as $x$ goes to infinity using $\epsilon-\delta$

Hi i need to prove that $$\lim_{x\to \infty}\frac{x-3}{x^2 +1}=0$$ using the formal definition of a limit. Can anyone help?

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If you divide both the numerator and the denominator by $x^2$, what do you get? What can you say about that expression? – Chris Feb 20 '13 at 2:18

Let $\,\epsilon>0\,$ , and since we're interested in $\,x\to\infty\,$ let us assume $\,x>3\,$ :

$$(**)\;\;\;\;\;\left|\frac{x-3}{x^2+1}\right|=\frac{x-3}{x^2+1}\leq \frac{x}{x^2}=\frac{1}{x}<\epsilon\Longleftrightarrow x>\frac{1}{\epsilon}$$

Thus, choosing $\,\delta=\frac{1}{\epsilon}\,$ , from the above we get that for all $\,x>\delta\,$ then (**) is true, which means

$$\lim_{x\to\infty}\frac{x-3}{x^2+1}=0$$

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thank you friends – PooperScooper Feb 20 '13 at 2:34

Hint: Use the fact $$0\leq\left|\frac{x-3}{x^2+1}\right|\leq\left|\frac{x}{x^2+1}\right|\leq\left|\frac{x}{x^2}\right|\leq\frac{1}{x}.$$ Can you show this right limit using $\varepsilon-\delta$ and conclude what the limit is?

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Hint: for $x\geq 3$ you have $$0\leq\frac{x-3}{x^2+1}\leq \frac{x}{x^2}=\frac{1}{x}.$$

Can you manage the $\epsilon$ proof now?

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