લક્ષની કિંમત શોધો: lim _{x rightarrow 0} frac | vedclass

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

Question

(B) ત્રિકોણમિતીય નિત્યસમ $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$ અને $\cos C + \cos D = 2 \cos \left(\fr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) ત્રિકોણમિતીય નિત્યસમ $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}ight) \sin \left(\frac{C-D}{2}ight)$ અને $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}ight) \cos \left(\frac{C-D}{2}ight)$ નો ઉપયોગ કરતા: અંશ: $4[\sin (2022 x) - \sin (2020 x)] = 4 	imes 2 \cos(2021 x) \sin(x) = 8 \cos(2021 x) \sin(x)$. છેદ: $x[\cos (2022 x) + \cos (2020 x) + 2 \cos (2021 x)] = x[2 \cos(2021 x) \cos(x) + 2 \cos(2021 x)] = 2x \cos(2021 x) [\cos(x) + 1]$. આ કિંમતો લક્ષમાં મૂકતા: $\lim _{x ightarrow 0} \frac{8 \cos(2021 x) \sin(x)}{2x \cos(2021 x) [\cos(x) + 1]} = \lim _{x ightarrow 0} \frac{4 \sin(x)}{x [\cos(x) + 1]}$. કારણ કે $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$ અને $\cos(0) = 1$: $= \frac{4(1)}{1 + 1} = \frac{4}{2} = 2$.

લક્ષની કિંમત શોધો: $\lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]}$

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}}}{{\sin {x^2}}} = $

$\mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right) = $

આપેલ લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{bx}$

$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} =$

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module