(c) $ \lim _ { x \
ightarrow 1 ^ { - } } \frac { | x - 1 | } { x ^ { 2 } + 6 x + 5 } $

Question

Accepted Answer

# Explanation:
## Step1: Analyze the absolute value
When $x\to1^{-}$, $x - 1<0$, so $|x - 1|=-(x - 1)$

## Step2: Factor the denominator
Factor $x^{2}+6x + 5=(x + 1)(x + 5)$

## Step3: Substitute and simplify
$$
\begin{align*}
\lim_{x\to1^{-}}\frac{|x - 1|}{x^{2}+6x + 5}&=\lim_{x\to1^{-}}\frac{-(x - 1)}{(x + 1)(x + 5)}\
&=\frac{-(1 - 1)}{(1+ 1)(1 + 5)}\
&=\frac{0}{2\times6}
\end{align*}
$$

# Answer:
$0$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the absolute value

Step2: Factor the denominator

Step3: Substitute and simplify

Answer: