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.

35. \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{{\cos x + {e^x} - 1}}\)

Short answer

The value of the limit is \(0\) .

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:

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

Thus, the value of the limit is 0.