Question

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.

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

Short answer

The value of the limit is \(\frac{1}{3}\) .

Step-by-step solution

Definition of l’Hospital’s Rule

Let \(f\) and \(g\) be differentiable and \(g'\left( x \right) \ne 0\) on an open interval I, containing \(a\) (except possibly at \(a\)). Assume that \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = 0\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = 0\), or \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \pm \infty \), and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = \pm \infty \).

(In certain, there is anindeterminate formof type \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\).) Then

\(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)

The limit exists on the right-hand side (or is \(\infty \) or \( - \infty \)).

Determine the limit

Evaluate the limit as shown below:

\(\mathop {\lim }\limits_{x \to 1} \frac{{x\sin x\left( {x - 1} \right)}}{{2{x^2} - x - 1}} = \frac{{\sin 1\left( {1 - 1} \right)}}{{2{{\left( 1 \right)}^2} - 1 - 1}}\)

The limit is in indeterminate form \(\frac{0}{0}\).

Apply l’Hospital Rule as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} \frac{{x\sin x\left( {x - 1} \right)}}{{2{x^2} - x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {x\sin x\left( {x - 1} \right)} \right)}}{{\frac{d}{{dx}}\left( {2{x^2} - x - 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{x\cos \left( {x - 1} \right) + \sin \left( {x - 1} \right)}}{{4x - 1}}\\ = \frac{{\cos \left( {1 - 1} \right) + \sin \left( {1 - 1} \right)}}{{4\left( 1 \right) - 1}}\\ = \frac{{\cos 0 + \sin 0}}{{4 - 1}}\\ = \frac{{1 + 0}}{3}\\ = \frac{1}{3}\end{array}\)

Thus, the value of the limit is \(\frac{1}{3}\).