mathop {lim }limits_{x to 0} frac{{sqrt {1 - | vedclass

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

Question

(C) આપણી પાસે $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) આપણી પાસે $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin^2(x^2/2)} }}{{2 \sin^2(x/2)}}$ છે.$x 	o 0$ માટે,$\sin(x^2/2) > 0$ હોવાથી:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt{2} \sin(x^2/2)}}{{2 \sin^2(x/2)}} = \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \frac{\sin(x^2/2)}{\sin^2(x/2)}$.$(x^2/2)$ અને $(x/2)^2$ વડે ગુણતા અને ભાગતા:$= \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{(\sin(x/2)/(x/2))^2 \cdot (x^2/4)} ight)$.$= \frac{1}{\sqrt{2}} \cdot \frac{1/2}{1/4} \cdot 1 = \sqrt{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ ની કિંમત શોધો.

Similar Questions

$\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{x - a}} = $

$\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2+x^3}$ ની કિંમત શોધો.

જો $f: R \rightarrow R$ એ $f(x) = [x-3] + |x-4|$ દ્વારા $x \in R$ માટે વ્યાખ્યાયિત હોય,તો $\lim_{x \rightarrow 3^{-}} f(x)$ ની કિંમત શોધો.

જો $\mathop {\lim }\limits_{n \to \infty } \frac{1 - 10^n}{1 + 10^{n+1}} = \frac{-\alpha}{10}$ હોય,તો $\alpha$ ની કિંમત શોધો.

$\lim\limits_{x \rightarrow 0}\left(\frac{3 x^{2}+2}{7 x^{2}+2}\right)^{\frac{1}{x^{2}}}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module