lim _{x rightarrow 0} frac{sin ^2 x}{sqrt{2}- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$Rationalizing the denominator:$= \lim _{x \rightarrow 0

Vedclass · Accepted Answer

$4 \sqrt{2}$

Vedclass · Answer

(B) We evaluate the limit: $\lim _{x ightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$Rationalizing the denominator:$= \lim _{x ightarrow 0} \frac{\sin ^2 x (\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}$$= \lim _{x ightarrow 0} \frac{\sin ^2 x (\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$Using the identity $\sin ^2 x = 1-\cos ^2 x = (1-\cos x)(1+\cos x)$:$= \lim _{x ightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$$= \lim _{x ightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$Substituting $x = 0$:$= (1+\cos 0)(\sqrt{2}+\sqrt{1+\cos 0})$$= (1+1)(\sqrt{2}+\sqrt{1+1})$$= 2 	imes (\sqrt{2}+\sqrt{2})$$= 2 	imes 2 \sqrt{2} = 4 \sqrt{2}$

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals

Similar Questions

The true statement for $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$ is

Let $f(x) = \frac{\ln(x^2 + e^x)}{\ln(x^4 + e^{2x})}$. If $\lim_{x \to \infty} f(x) = l$ and $\lim_{x \to -\infty} f(x) = m$,then:

If $0 < p < q$,then $\lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}$ is equal to :

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\pi - x}$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^5} - 1}}{{{{(1 + x)}^3} - 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module