Answer to Question #244119 in Calculus for CJNS

Question

Evaluate the limit. Lim (3x^3-4x+2)/(w^3-5). x-->2

Answer 1

We substitute x = 2 on the expresison and calculate the limit as:

$\lim\limits_{x\, \to \,2} \cfrac{3x^3-4x+2}{x^3-5}=\cfrac{3(2)^3-4(2)+2}{(2)^3-5} \\ \lim\limits_{x\, \to \,2} \cfrac{3x^3-4x+2}{x^3-5}=\cfrac{24-8+2}{8-5} \\ \lim\limits_{x\, \to \,2} \cfrac{3x^3-4x+2}{x^3-5}=\cfrac{18}{3}=6$

In conclusion, the limit is equal to 6.

Reference:

• Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.