જો a 0 અને b

Question

(A) આપેલ લક્ષ: $L = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$નિત્યસમ $1 - \cos 2	heta = 2\sin^2 	heta$ નો ઉપ

Vedclass · Accepted Answer

$\frac{a\sqrt{2}}{b}$

Vedclass · Answer

(A) આપેલ લક્ષ: $L = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$નિત્યસમ $1 - \cos 2	heta = 2\sin^2 	heta$ નો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt {2\sin^2 ax} }}{{\sin bx}} = \mathop {\lim }\limits_{x 	o {0^ + }} \frac{{\sqrt{2} |\sin ax|}}{{\sin bx}}$અહીં $x 	o 0^+$ અને $a > 0$ હોવાથી,$ax > 0$,તેથી $|\sin ax| = \sin ax$.$L = \sqrt{2} \mathop {\lim }\limits_{x 	o {0^ + }} \frac{\sin ax}{\sin bx} = \sqrt{2} \mathop {\lim }\limits_{x 	o {0^ + }} \left( \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx} ight)$$L = \sqrt{2} \cdot 1 \cdot 1 \cdot \frac{a}{b} = \frac{a\sqrt{2}}{b}$

જો $a > 0$ અને $b < 0$ હોય,તો $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$ ની કિંમત શોધો.

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $

આપેલ લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx}$,જ્યાં $a, b \neq 0$.

$\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos 6x}{x} = $

$\lim _{x}$ ${\rightarrow 0} \left( \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \right) \left( \frac{\sin ^3(4 x)}{(\log _e(2 x+1))^5} \right) \right)$ ની કિંમત $.........$ છે.

$\mathop {\lim }\limits_{x \to 0} \frac{\sin 2x}{x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module