Question

find the limit.
lim_{x
ightarrow0}\frac{\frac{1}{x - 1}+\frac{1}{x + 1}}{4x}

Explanation:

Step1: Combine the fractions in the numerator

First, find a common - denominator for $\frac{1}{x - 1}+\frac{1}{x + 1}$. The common denominator is $(x - 1)(x + 1)=x^{2}-1$. So, $\frac{1}{x - 1}+\frac{1}{x + 1}=\frac{x + 1+(x - 1)}{x^{2}-1}=\frac{2x}{x^{2}-1}$.

Step2: Rewrite the original limit

The original limit $\lim_{x
ightarrow0}\frac{\frac{1}{x - 1}+\frac{1}{x + 1}}{4x}$ becomes $\lim_{x
ightarrow0}\frac{\frac{2x}{x^{2}-1}}{4x}=\lim_{x
ightarrow0}\frac{2x}{4x(x^{2}-1)}$.

Step3: Simplify the fraction

Cancel out the common factor $2x$ (since $x
eq0$ when taking the limit) in the numerator and denominator. We get $\lim_{x
ightarrow0}\frac{1}{2(x^{2}-1)}$.

Step4: Evaluate the limit

Substitute $x = 0$ into $\frac{1}{2(x^{2}-1)}$. We have $\frac{1}{2(0^{2}-1)}=-\frac{1}{2}$.

Answer:

$-\frac{1}{2}$