$\displaystyle\lim_{x\to0} \frac{\sin^{2}x}{\sqrt{2} - \sqrt{1+\cos x}} $ equals :

Question

If $\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1}-1) \log_e(x-1)} = m$ (exists finitely), then find the value of $a + b + m$.

Answer 1

To solve the problem, we need to evaluate the limit: $\displaystyle\lim_{x\to0} \frac{\sin^{2}x}{\sqrt{2} - \sqrt{1+\cos x}}$.

Start by using the limit property and the approximation for small $x$: $\cos x \approx 1 - \frac{x^2}{2}$ and $\sin x \approx x$ as $x \to 0$.
Substitute these approximations into the expression: $\sin^2 x \approx x^2$ and $\cos x \approx 1 - \frac{x^2}{2}$.
Rewrite the denominator using the approximation for the square root: \[ \sqrt{1 + \cos x} \approx \sqrt{2 - \frac{x^2}{2}} \approx \sqrt{2}\left(1 - \frac{x^2}{8}\right) \] Thus, \[ \sqrt{2} - \sqrt{1 + \cos x} \approx \sqrt{2} - \sqrt{2}\left(1 - \frac{x^2}{8}\right) = \sqrt{2}\left(\frac{x^2}{8}\right) \]
Now, substitute back into the original limit expression: $\displaystyle\lim_{x\to0} \frac{x^2}{\sqrt{2} \cdot \frac{x^2}{8}}$.
Simplify the expression: \[ \frac{x^2}{\sqrt{2} \cdot \frac{x^2}{8}} = \frac{x^2 \cdot 8}{x^2 \cdot \sqrt{2}} = \frac{8}{\sqrt{2}} \]
Rationalize the denominator: \[ \frac{8}{\sqrt{2}} = \frac{8 \times \sqrt{2}}{2} = 4\sqrt{2} \]
Therefore, the evaluated limit is $4\sqrt{2}$.

Hence, the correct answer is $4 \sqrt{2}$.

Answer 2

To solve the problem, we need to evaluate the limit: $\displaystyle\lim_{x\to0} \frac{\sin^{2}x}{\sqrt{2} - \sqrt{1+\cos x}}$.

Start by using the limit property and the approximation for small $x$: $\cos x \approx 1 - \frac{x^2}{2}$ and $\sin x \approx x$ as $x \to 0$.
Substitute these approximations into the expression: $\sin^2 x \approx x^2$ and $\cos x \approx 1 - \frac{x^2}{2}$.
Rewrite the denominator using the approximation for the square root: \[ \sqrt{1 + \cos x} \approx \sqrt{2 - \frac{x^2}{2}} \approx \sqrt{2}\left(1 - \frac{x^2}{8}\right) \] Thus, \[ \sqrt{2} - \sqrt{1 + \cos x} \approx \sqrt{2} - \sqrt{2}\left(1 - \frac{x^2}{8}\right) = \sqrt{2}\left(\frac{x^2}{8}\right) \]
Now, substitute back into the original limit expression: $\displaystyle\lim_{x\to0} \frac{x^2}{\sqrt{2} \cdot \frac{x^2}{8}}$.
Simplify the expression: \[ \frac{x^2}{\sqrt{2} \cdot \frac{x^2}{8}} = \frac{x^2 \cdot 8}{x^2 \cdot \sqrt{2}} = \frac{8}{\sqrt{2}} \]
Rationalize the denominator: \[ \frac{8}{\sqrt{2}} = \frac{8 \times \sqrt{2}}{2} = 4\sqrt{2} \]
Therefore, the evaluated limit is $4\sqrt{2}$.

Hence, the correct answer is $4 \sqrt{2}$.

The Correct Option is B