Determine the value of the limit $\lim _ { x \rightarrow 0 } \frac { 1 - \cos x } { x \sin x }$ ?
A) $- \infty$
B) $- \frac { 1 } { 2 }$
C) $\frac { 1 } { 2 }$
D)0
E) $\infty$

Question

Quizplus · Accepted Answer

The limit can be evaluated using L'Hôpital's Rule, which allows us to differentiate the numerator and the denominator separately because the original limit is of the indeterminate form 0/0. Differentiating the numerator gives $ \sin x $, and differentiating the denominator gives $ \sin x + x \cos x $. Applying L'Hôpital's Rule, we get $\lim _ { x \rightarrow 0 } \frac { \sin x } { \sin x + x \cos x }$. Evaluating this limit as $x$ approaches 0, we find that both the numerator and the denominator approach 0, but using L'Hôpital's Rule again or simply substituting $x = 0$ after the first differentiation (since the direct substitution now gives a determinate form), we find the limit to be $\frac{1}{2}$, as the terms involving $x$ in the denominator become negligible, leaving us with $\frac{\sin x}{\sin x}$ which simplifies to 1, and considering the original expression's structure, the correct answer is indeed $\frac{1}{2}$.

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Determine the Value of the Limit $\lim _ { x \rightarrow 0 } \frac { 1 - \cos x } { x \sin x }$