lim_{x\\to - 1}\\frac{x^{2}+4}{x + 2}=
lim_{x\\to - 1}\\frac{x^{2}+4}{x + 2}=
Answer
Explanation:
Step1: Substitute x value
Substitute $x = - 1$ into $\frac{x^{2}+4}{x + 2}$. $\frac{(-1)^{2}+4}{-1 + 2}$
Step2: Simplify the expression
First, calculate the numerator $(-1)^{2}+4=1 + 4=5$, and the denominator $-1 + 2=1$. $\frac{5}{1}=5$
Answer:
$5$