Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
If you divide both the numerator and the denominator by $x^2$, what do you get? What can you say about that expression? –  user1296727 Feb 20 '13 at 2:18
add comment

3 Answers 3

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


share|improve this answer
thank you friends –  PooperScooper Feb 20 '13 at 2:34
add comment

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?

share|improve this answer
add comment

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?

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.