Answer
$32$
Work Step by Step
Given that: $\lim\limits_{x \to 2}\frac{x^4-16}{x-2}$
let's factorize the numerator:
\[ x^4 - 16 = (x^2 + 4)(x^2 - 4)\]\[=(x^2 + 4)(x + 2)(x - 2)\]
Now, we can simplify the expression:
\[ \lim_{x \to 2} \frac{x^4 - 16}{x - 2} = \lim_{x \to 2} \frac{(x^2 + 4)(x + 2)(x - 2)}{x - 2} \]\[ = \lim_{x \to 2} (x^2 + 4)(x + 2) \]Now, we can directly substitute \( x = 2 \) into the expression:
\[=(2^2 + 4)(2 + 2)\]\[=(4 + 4)(4)\]\[=(8)(4)\]\[ = 32 \]
So, $\lim\limits_{x \to 2}\frac{x^4-16}{x-2}= 32$.
