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.

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

Short answer

The value of \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}\) is \(2\).

Step-by-step solution

Given information

The given limit is \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}\).

Concept of L'Hospital's Rule

Suppose\(f\)and\(g\)are differentiable and\({g^\prime }(x) \ne 0\)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\( - \infty \)).

Find the value of the limit

Obtain the value of the function as\(x\)approaches\(0\).

As\(x\)approaches\(0\), the numerator becomes:

\(\begin{array}{c}{e^x} - {e^{ - x}} - 2x = {e^0} - {e^{ - 0}} - 2 \cdot (0)\\ = 1 - 1 - 0\\ = 0\end{array}\)

And the denominator becomes:

\(\begin{array}{c}x - \sin x = 0 - \sin 0\\ = 0 - 0\\ = 0\end{array}\)

Thus,\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}} = \frac{0}{0}\)is in an indeterminate form.

Therefore, apply L'Hospital's Rule and obtain the limit.

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

Again as\(x\)approaches\(0\), the numerator is:

\(\begin{array}{c}{e^x} + {e^{ - x}} - 2 = {e^0} + {e^{ - 0}} - 2\\ = 1 + 1 - 2\\ = 2 - 2\\ = 0\end{array}\)

And as\(x\)approaches\(0\), the denominator is:

\(\begin{array}{c}1 - \cos x = 1 - \cos 0\\ = 1 - 1\\ = 0\end{array}\)

Thus, \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{1 - \cos x}} = \frac{0}{0}\) is again in an indeterminate form.

Apply L'Hospital's Rule and obtain the limit

Therefore, apply L'Hospital's Rule again and obtain the limit.

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}\)

Again as\(x\)approaches\(0\), the numerator is:

\(\begin{array}{c}{e^x} - {e^{ - x}} = {e^0} - {e^{ - 0}}\\ = 1 - 1\\ = 0\end{array}\)

And as\(x\)approaches\(0\), the denominator is:

\(\begin{array}{c}\sin x = \sin 0\\ = 0\end{array}\)

Thus,\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\sin x}}\)is once again in an indeterminate form.

Apply L'Hospital's Rule again and obtain the limit.

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

Thus, the limit of the function is \(2\).