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.

• You can expand the denominator and use polynomial division. – Ataraxia May 25 '13 at 20:12
• Hint, Simplify the denominator and then divide everything by the the remaining $3x^2$ term. Got it? – Amzoti May 25 '13 at 20:15
• 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. – Thums May 25 '13 at 20:19
• @LuanCristianThums: I have those ALL the time! :-) You are very welcome! – Amzoti May 25 '13 at 20:19

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 $\frac{1}{3}$.

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.

• Thank you for your answer, helped me! – Thums May 25 '13 at 20:19

You can expand the denominator and use polynomial division:

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

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