Question

1–38 ■ Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

1. \(\mathop {lim}\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^2} - x}}\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^2} - x}}\) is \(2\).

Step-by-step solution

Given limit

The given limit is \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^2} - x}}\).

The Concept of L'Hospital's Rule

L'Hospital's Rule:

Suppose\(f\)and\(g\)are differentiable and\(g'(x) \ne 0\) near\(a\)(except possibly at\(a\)).

Suppose that\(\mathop {\lim }\limits_{x \to a} f(x) = 0\)and\(\mathop {\lim }\limits_{x \to a} g(x) = 0\)or that\(\mathop {lim}\limits_{x \to a} f(x) = \pm \infty\) and \(\mathop {\lim }\limits_{x \to a} g(x) = \pm \infty \)or in other words, we have an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\),

Then, \(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\)if the limit on the right side exists (or is\(\infty \) or\( - \infty \)).

Evaluate the given limit

Simplify the given limit as shown below.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^2} - x}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x + 1)(x - 1)}}{{x(x - 1)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{(x + 1)}}{x}\\ = \frac{{1 + 1}}{1}\\ = 2\end{array}\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^2} - x}}\) is \(2\).