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