Question

How do you find the limit of `(-2x^3)/(9-x^2)` as x approaches -3 from the left?

Answer 1

`-\infty`

Explanation:

Given that

`\lim_{x\to -3^{-}}\frac{-2x^3}{9-x^2}`

`=\lim_{x\to -3^{-}}\frac{-2x^3}{(3-x)(3+x)}`

`=\lim_{x\to -3^{-}}\frac{-2x^3}{3-x}\cdot\frac{1}{3+x}`

Notice `\lim_{x\to -3^{-}}\frac{-2x^3}{3-x}\to 9\ \ & \ \ \lim_{x\to -3^{-}}\frac{1}{3+x}\to -\infty`

`=-\infty`