Question

The value of $\lim_{x \to 0} \dfrac{\sqrt{1 - \cos(x^2)}}{1 - \cos x}$ is equal to:

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $2\sqrt{2}$
• E. $0$

Hint:

Use standard limits: $1-\cos x \approx \frac{x^2}{2}$ for small $x$.

Answer 1

The Correct Option is B

Concept:
• $1 - \cos x \approx \frac{x^2}{2}$ as $x \to 0$

Step 1: Apply approximation
\[ 1 - \cos(x^2) \approx \frac{x^4}{2} \] \[ \sqrt{1 - \cos(x^2)} \approx \sqrt{\frac{x^4}{2}} = \frac{x^2}{\sqrt{2}} \]

Step 2: Denominator
\[ 1 - \cos x \approx \frac{x^2}{2} \]

Step 3: Compute limit
\[ \frac{\frac{x^2}{\sqrt{2}}}{\frac{x^2}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \] Final Conclusion:
\[ = \sqrt{2} \]