lim_{x\\to - 4}\\frac{x^{2}-16}{x + 4}=-8
lim_{x\\to - 4}\\frac{x^{2}-16}{x + 4}=-8
Answer
Explanation:
Step1: Factor the numerator
We know that $x^{2}-16=(x + 4)(x - 4)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ where $a=x$ and $b = 4$. So the limit becomes $\lim_{x\rightarrow - 4}\frac{(x + 4)(x - 4)}{x + 4}$.
Step2: Simplify the function
Cancel out the common factor $(x + 4)$ in the numerator and the denominator. We get $\lim_{x\rightarrow - 4}(x - 4)$.
Step3: Evaluate the limit
Substitute $x=-4$ into the function $x - 4$. We have $-4-4=-8$.
Answer:
$-8$