Questions: Evaluate lim as x approaches 0 of (sin x)/(cos x-1)

Solution Steps

To evaluate the limit \(\lim _{x \rightarrow 0} \frac{\sin x}{\cos x-1}\), we can use L'Hôpital's Rule, which is applicable when the limit results in an indeterminate form like \(\frac{0}{0}\). By differentiating the numerator and the denominator separately, we can find the limit.

We need to evaluate the limit \( \lim _{x \rightarrow 0} \frac{\sin x}{\cos x - 1} \). As \( x \) approaches \( 0 \), both the numerator \( \sin x \) and the denominator \( \cos x - 1 \) approach \( 0 \), resulting in the indeterminate form \( \frac{0}{0} \).

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then: \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \] We differentiate the numerator and the denominator:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x - 1 \) is \( -\sin x \).

Now we evaluate the new limit: \[ \lim_{x \rightarrow 0} \frac{\cos x}{-\sin x} \] Substituting \( x = 0 \): \[ \frac{\cos(0)}{-\sin(0)} = \frac{1}{0} \] This indicates that the limit approaches \( -\infty \).

Final Answer