finding the $\lim_{x \rightarrow -2 }\frac{x+2}{x^3+8}$

Question

How do you solve $\lim_{x \rightarrow -2} \frac{x+2}{x^3+8}$ to get 1/12? I tried factoring out the denominator into $(x+2)(x+2)(x+2)$ and cancelling it out with the top but when you plug in 0 for x the denominator is still 0.

the bottom factors as $(x+2)(x^2-2x+4)$. – user84413 Oct 01 '14 at 00:24 — user84413, Oct 01 '14 at 00:24
Generally, $a^n+b^n\neq(a+b)^n$. – Lucian Oct 01 '14 at 00:55 — Lucian, Oct 01 '14 at 00:55

BlackAdder · Answer 1 · 2014-10-01T00:32:04.350

2

HINT: Notice that your factorisation is wrong.

$$x^3+8=(x+2)(x^2 −2x+4)$$

So, when $x=-2$, $\quad x^2 −2x+4=12$.
(You should not be plugging $0$ for $x$.)

edited Oct 01 '14 at 00:32

answered Oct 01 '14 at 00:26

BlackAdder

4,029

I meant -2 but I got it now – Jessica Oct 01 '14 at 00:56
@Jessica That's great! If you are happy with the answer, could you accept it please? Thanks! – BlackAdder Oct 01 '14 at 01:01

score 0 · Accepted Answer · answered Oct 01 '14 at 00:56

0

Here are the steps $$ \lim_{x\to -2} \frac{x+2}{x^3+8}=\lim_{x\to -2} \frac{x+2}{(x+2)(x^2-2x+4)} $$ $$ =\lim_{x\to -2} \frac{1}{x^2-2x+4}= \frac{1}{(-2)^2-2(-2)+4}=\frac{1}{4+4+4}=\frac{1}{12} $$

answered Oct 01 '14 at 00:56

k170

9,045

finding the $\lim_{x \rightarrow -2 }\frac{x+2}{x^3+8}$

2 Answers2