Question

Lim x tends to 2 ( x^3 - 8/x -2 )

Answer 1

This should be your answer.

Answer 2

We have:

$\lim_{x \to 2} \dfrac{x^3-8}{x-2}$

We have to find, the value of $\lim_{x \to 2} \dfrac{x^3-8}{x-2}$=?

Solution:

∴ $\lim_{x \to 2} \dfrac{x^3-8}{x-2}$  $(\dfrac{0}{0} form)$

= $\lim_{x \to 2} \dfrac{x^3-2^3}{x-2}$

Using the limit identity,

 $\lim_{x \to a} \dfrac{x^n-a^n}{x-a}=na^{n-1}$

Here, n = 3 and a = 2

= $3(2)^{3-1}$

= $3(2)^{2}$

= 3 × 4

= 12

∴ $\lim_{x \to 2} \dfrac{x^3-8}{x-2}$  = 12

Thus, the value of $\lim_{x \to 2} \dfrac{x^3-8}{x-2}$  is "equal to 12".