Chapter B, Problem 1E

To determine

To evaluate: The limit by using l’Hôpital’s Rule.

Answer to Problem 1E

The value of the limit is, $\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}=\frac{1}{2}$.

Explanation of Solution

l’Hôpital’s Rule:

Suppose f and g are differentiable on an open interval I that contains a, with possible exception of a itself and $g^{\prime }\left(x\right)\ne 0$ for all x in I, with possible exception of a. if $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ has an indeterminate form of the type $\frac{0}{0}\text{or }\frac{\infty }{\infty }$, then $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$ provided that the limit on the right or is infinite.

Calculation:

Obtain the value of limit $\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}$,

Since $\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}=\frac{0}{0}$ in an indeterminate form, apply the l’Hôpital’s Rule,

$\begin{array}{c}\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}=\underset{x\to 1}{\lim }\frac{\frac{d}{dx}\left(x-1\right)}{\frac{d}{dx}\left(x^2-1\right)}\\ =\underset{x\to 1}{\lim }\frac{1}{2x}\\ =\frac{1}{2\left(1\right)}\\ =\frac{1}{2}\end{array}$

Thus, the value of the limit is, $\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}=\frac{1}{2}$.