Question

c) $lim_{x
ightarrow5^{-}}\frac{x^{2}-25}{|2x - 10|}$

Explanation:

Step1: Factor the numerator

$x^{2}-25=(x + 5)(x - 5)$

Step2: Analyze the absolute - value for $x\to5^{-}$

When $x\to5^{-}$, $2x-10<0$, so $|2x - 10|=-(2x - 10)=10 - 2x$.

Step3: Rewrite the limit

$\lim_{x\to5^{-}}\frac{x^{2}-25}{|2x - 10|}=\lim_{x\to5^{-}}\frac{(x + 5)(x - 5)}{10 - 2x}$

Step4: Simplify the expression

$\lim_{x\to5^{-}}\frac{(x + 5)(x - 5)}{-2(x - 5)}=\lim_{x\to5^{-}}\frac{x + 5}{-2}$

Step5: Evaluate the limit

Substitute $x = 5$ into $\frac{x + 5}{-2}$, we get $\frac{5+5}{-2}=-5$

Answer:

$-5$