Question

d) $lim_{x \to 2}left(\frac{\frac{1}{x - 1}-\frac{3}{x + 1}}{x - 2}
ight)$

Explanation:

Step1: Combine the fractions in the numerator

First, find a common - denominator for $\frac{1}{x - 1}-\frac{3}{x + 1}$. The common denominator is $(x - 1)(x + 1)$. So, $\frac{1}{x - 1}-\frac{3}{x + 1}=\frac{(x + 1)-3(x - 1)}{(x - 1)(x + 1)}=\frac{x + 1-3x + 3}{(x - 1)(x + 1)}=\frac{-2x + 4}{(x - 1)(x + 1)}$.
The original limit becomes $\lim_{x
ightarrow2}\frac{\frac{-2x + 4}{(x - 1)(x + 1)}}{x - 2}=\lim_{x
ightarrow2}\frac{-2x + 4}{(x - 1)(x + 1)(x - 2)}$.

Step2: Factor the numerator

Factor out - 2 from the numerator: $-2x + 4=-2(x - 2)$.
The limit is now $\lim_{x
ightarrow2}\frac{-2(x - 2)}{(x - 1)(x + 1)(x - 2)}$.

Step3: Cancel out the common factor

Cancel out the common factor $(x - 2)$ (since $x
eq2$ when taking the limit).
We get $\lim_{x
ightarrow2}\frac{-2}{(x - 1)(x + 1)}$.

Step4: Substitute $x = 2$

Substitute $x = 2$ into $\frac{-2}{(x - 1)(x + 1)}$.
$\frac{-2}{(2 - 1)(2 + 1)}=\frac{-2}{1\times3}=-\frac{2}{3}$.

Answer:

$-\frac{2}{3}$