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

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

Question

(C) ત્રિકોણમિતીય નિત્યસમ $1 - \cos 	heta = 2 \sin^2(	heta/2)$ નો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{1 - \cos nx} = \mat

Vedclass · Accepted Answer

$\frac{m^2}{n^2}$

Vedclass · Answer

(C) ત્રિકોણમિતીય નિત્યસમ $1 - \cos 	heta = 2 \sin^2(	heta/2)$ નો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{1 - \cos nx} = \mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2(mx/2)}{2 \sin^2(nx/2)}$$= \mathop {\lim }\limits_{x 	o 0} \left[ \left( \frac{\sin(mx/2)}{mx/2} ight)^2 \cdot \frac{m^2 x^2}{4} \cdot \left( \frac{nx/2}{\sin(nx/2)} ight)^2 \cdot \frac{4}{n^2 x^2} ight]$કારણ કે $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,તેથી:$= \frac{m^2}{n^2} \cdot 1 = \frac{m^2}{n^2}$.વૈકલ્પિક રીત: $L$-Hospital નો નિયમ વાપરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{1 - \cos nx} = \mathop {\lim }\limits_{x 	o 0} \frac{m \sin mx}{n \sin nx}$$= \mathop {\lim }\limits_{x 	o 0} \frac{m^2 \cos mx}{n^2 \cos nx} = \frac{m^2(1)}{n^2(1)} = \frac{m^2}{n^2}$.

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

Similar Questions

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

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ ની કિંમત શોધો.

લક્ષ $\lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,\cot \,\left( {4x} \right)}}{{{{\sin }^2}\,x\,{{\cot }^2}\,\left( {2x} \right)}}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module