13) $lim_{x\rightarrow - 5^{-}}\frac{2}{x^{2}-25}$
13) $lim_{x\rightarrow - 5^{-}}\frac{2}{x^{2}-25}$
Answer
Explanation:
Step1: Analyze the numerator of the expression.
The numerator is a constant value, $2$.
Step2: Analyze the denominator as $x$ approaches $-5$ from the left side.
We need to evaluate the behavior of $x^2 - 25$ as $x \to -5^-$. Let $x$ be a number slightly less than $-5$. For example, $x = -5.1$. Then $x^2 = (-5.1)^2 = 26.01$. So, $x^2 - 25 = 26.01 - 25 = 1.01$. As $x$ approaches $-5$ from the left side (e.g., $x = -5.01, -5.001, \dots$), $x^2$ will be slightly greater than $25$. Therefore, $x^2 - 25$ will be a small positive number. We can denote this as $0^+$.
Step3: Determine the limit of the fraction.
We are evaluating the limit of the form $\frac{2}{0^+}$. When the numerator is a positive constant and the denominator approaches $0$ from the positive side, the limit tends to positive infinity. $$ \lim_{x \to -5^-} \frac{2}{x^2 - 25} = \frac{2}{0^+} = +\infty $$
Answer:
$$ +\infty $$