Question

$\lim_{x \to 0} \frac{\sqrt[3]{\cos x} - \sqrt{\cos x}}{\sin^2 x} =$

• A. 1/2
• B. 1/12
• C. 1/6
• D. 2/3

Hint:

For limits involving trigonometric functions as $x \to 0$, using standard approximations like $\sin x \approx x$, $\cos x \approx 1 - x^2/2$, and the binomial approximation $(1+u)^n \approx 1+nu$ is often faster and less error-prone than repeated applications of L'Hopital's Rule.

Answer 1

The Correct Option is B

Step 1: Use standard limits and algebraic manipulation.
The limit is of the indeterminate form $0/0$. We can use the standard limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$. This implies that for small $x$, $\sin^2 x \approx x^2$. The limit becomes: \[ L = \lim_{x \to 0} \frac{(\cos x)^{1/3} - (\cos x)^{1/2}}{x^2}. \] We can factor out $(\cos x)^{1/2}$: \[ L = \lim_{x \to 0} \frac{(\cos x)^{1/2} [(\cos x)^{-1/6} - 1]}{x^2}. \] As $x \to 0$, $(\cos x)^{1/2} \to 1$. So, \[ L = \lim_{x \to 0} \frac{(\cos x)^{-1/6} - 1}{x^2}. \]

Step 2: Use Taylor expansion for $\cos x$.
For small $x$, the Taylor expansion of $\cos x$ is $1 - \frac{x^2}{2!} + O(x^4)$. \[ L = \lim_{x \to 0} \frac{(1 - x^2/2)^{-1/6} - 1}{x^2}. \]

Step 3: Use the binomial approximation.
The binomial approximation for $(1+u)^n$ for small $u$ is $1+nu$. Here, $u = -x^2/2$ and $n = -1/6$. \[ (1 - x^2/2)^{-1/6} \approx 1 + (-\frac{1}{6})(-\frac{x^2}{2}) = 1 + \frac{x^2}{12}. \] Substitute this back into the limit: \[ L = \lim_{x \to 0} \frac{(1 + x^2/12) - 1}{x^2} = \lim_{x \to 0} \frac{x^2/12}{x^2}. \]

Step 4: Evaluate the final limit.
\[ L = \frac{1}{12}. \] \[ \boxed{L = \frac{1}{12}}. \]