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.

38. \(\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\sin x}}{{\sin x - x}}\)

Short answer

The value of the limit is \( - 6\).

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:

\(\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\sin x}}{{\sin x - x}} = \frac{{{0^2}\sin 0}}{{\sin 0 - 0}}\)

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

Apply l’Hospital Rule as shown below:

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

It is observed that the limit is in indeterminate form \(\frac{0}{0}\).

Apply l’Hospital Rule as shown below:

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

It is observed that the limit is in indeterminate form \(\frac{0}{0}\).

Apply l’Hospital Rule as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - {x^2}} \right)\sin x + 4x\cos x}}{{ - \sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left( {\left( {2 - {x^2}} \right)\sin x + 4x\cos x} \right)}}{{\frac{d}{{dx}}\left( { - \sin x} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - {x^2}} \right)\cos x - 2x\sin x - 4x\sin x + 4\cos x}}{{ - \cos x}}\\ = \frac{{\left( {2 - {0^2}} \right)\cos 0 - 2\left( 0 \right)\sin 0 - 4\left( 0 \right)\sin 0 + 4\cos 0}}{{ - \cos 0}}\\ = \frac{{2\left( 1 \right) - 0 - 0 + 4\left( 1 \right)}}{{ - 1}}\\ = - 6\end{array}\)

Thus, the value of the limit is \( - 6\).