lim_{h rightarrow 0} 2 left{ frac{sqrt{3} sin | vedclass

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

Question

(A) ધારો કે $L = \lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \s

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) ધારો કે $L = \lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$.$\sin(A+B) = \sin A \cos B + \cos A \sin B$ અને $\cos(A+B) = \cos A \cos B - \sin A \sin B$ ના વિસ્તરણનો ઉપયોગ કરતા:અંશ $= \sqrt{3} (\sin \frac{\pi}{6} \cos h + \cos \frac{\pi}{6} \sin h) - (\cos \frac{\pi}{6} \cos h - \sin \frac{\pi}{6} \sin h)$$= \sqrt{3} (\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h) - (\frac{\sqrt{3}}{2} \cos h - \frac{1}{2} \sin h)$$= \frac{\sqrt{3}}{2} \cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2} \cos h + \frac{1}{2} \sin h = 2 \sin h$.લિમિટમાં કિંમત મૂકતા:$L = \lim_{h \rightarrow 0} 2 \left( \frac{2 \sin h}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right) = \frac{4}{\sqrt{3}} \lim_{h \rightarrow 0} \left( \frac{\sin h}{h} \right) \cdot \lim_{h \rightarrow 0} \left( \frac{1}{\sqrt{3} \cos h - \sin h} \right)$.$\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$ અને $\lim_{h \rightarrow 0} (\sqrt{3} \cos h - \sin h) = \sqrt{3}(1) - 0 = \sqrt{3}$ હોવાથી:$L = \frac{4}{\sqrt{3}} \cdot 1 \cdot \frac{1}{\sqrt{3}} = \frac{4}{3}$.

$\lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$ ની કિંમત શોધો.

Similar Questions

જો $\lim _{x \rightarrow 0}\left(\frac{1+cx}{1-cx}\right)^{1/x}=4$ હોય, તો $\lim _{x \rightarrow 0}\left(\frac{1+2cx}{1-2cx}\right)^{1/x}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} \right) - \frac{\pi }{4}} \right]$ ની કિંમત શોધો.

$[x]$ એ $x$ થી નાનો અથવા તેના જેટલો મહત્તમ પૂર્ણાંક દર્શાવે છે. જો $\{x\}=x-[x]$ અને $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$ હોય,તો $\sin \theta+\cos \theta=$

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module