1
$\begingroup$

I'm doing some exercises of limits approaching infinite, most are simple polynomials where only the highest degree term will matter in the end but for this one I couldn't find a solution (not correct at least).

$$\lim_{x\to-\infty}\frac{x^2+x+1}{(x+1)^3-x^3}$$

How should I proceed to get the correct answer? ($\frac{1}{3}$)

Also, while simplifying some questions this question took my mind: is it correct to say that $\sqrt{a+b}$ is equal to $\sqrt{a}+\sqrt{b}$?

Ps.: Without using l'Hôpital's rule.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ You can expand the denominator and use polynomial division. $\endgroup$ – Ataraxia May 25 '13 at 20:12
  • 1
    $\begingroup$ Hint, Simplify the denominator and then divide everything by the the remaining $3x^2$ term. Got it? $\endgroup$ – Amzoti May 25 '13 at 20:15
  • $\begingroup$ Yes @Amzoti, got it. Thank you both, as it seems I had a brain malfunction and was considering $(x+1)^3$ = $x^3+1^3$. Thank you so much. $\endgroup$ – Thums May 25 '13 at 20:19
  • $\begingroup$ @LuanCristianThums: I have those ALL the time! :-) You are very welcome! $\endgroup$ – Amzoti May 25 '13 at 20:19
1
$\begingroup$

Start by expanding:

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

Simplify to obtain:

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

Now divide every term by $x^{2}$:

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

Now take the limit to obtain $\frac{1}{3}$.

Improve this answer
$\endgroup$
1
$\begingroup$

In general $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$ (e.g. $\sqrt{2} \neq \sqrt{1} + \sqrt{1} = 2$).

The first problem can be solved by writing down what $(x+1)^3$ means in terms of a polynomial.

Improve this answer
$\endgroup$
  • $\begingroup$ Thank you for your answer, helped me! $\endgroup$ – Thums May 25 '13 at 20:19
1
$\begingroup$

You can expand the denominator and use polynomial division:

$$(x+1)^3-x^3=3x^2+3x+1$$

$$\frac{x^2+x+1}{3x^2+3x+1}=\frac{1}{3}-\frac{2}{3}\frac{1}{3x^2+3x+1}$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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