Question

Find the Limit:

Answer 1

\[\lim_{x \rightarrow -2} \frac{x+2}{x^3+8}\]

Answer 2

`Sum of Cubes Formula`:\[\Large a^3+b^3 \qquad=\qquad (a+b)(a^2-ab+b^2)\]That will allow us to change our denominator a bit :)

Answer 3

\[\Large x^3+8\qquad=\qquad x^3+2^3\qquad=\qquad (x+2)(x^2-2x+2^2)\]Understand how that works? :o

Answer 4

Haha, yes @zepdrix :)
Couldn't we also use L'hoital's rule?

Answer 5

Since we're getting the indeterminate form 0/0, yes L'Hop would work just fine! :)

Answer 6

\[\lim_{X \rightarrow -2}\frac{x+2}{x^3+8}=\lim_{X \rightarrow -2}\frac{x+2}{x^3+2^3}\]
\[=\lim_{X \rightarrow -2}\frac{x+2}{(x+2)(x^2-2x+2^2}=\lim_{X \rightarrow -2}\frac{x+2}{(x+2)(x^2-2x+4}\]
\[=\lim_{X \rightarrow -2}\frac{1}{(x^2-2x+4}\]
Applying limit
\[=\frac{1}{(-2)^2-2(-2)+4} =\frac{1}{4+4+4} =\frac{1}{12} \]

Answer 7

Thanks guys :)
@zepdrix @dpasingh