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

1 Answer

Harish Chandra Rajpoot

Jul 23, 2018

#-\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#

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