lim _{x rightarrow 0} frac{(1-cos 2 x)(3+cos | vedclass

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Question

(A) આપણે જાણીએ છીએ કે $1 - \cos 2x = 2 \sin^2 x$. આ કિંમત લક્ષમાં મૂકતા: $\lim _{x ightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x 	an 4x}$ $= \lim _

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $1 - \cos 2x = 2 \sin^2 x$. આ કિંમત લક્ષમાં મૂકતા: $\lim _{x ightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x 	an 4x}$ $= \lim _{x}$ ${ightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{x}{	an 4x} \cdot (3 + \cos x) ight]$ $= \lim _{x}$ ${ightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{1}{4 \cdot \frac{	an 4x}{4x}} \cdot (3 + \cos x) ight]$ પ્રમાણિત લક્ષ $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$ અને $\lim_{	heta ightarrow 0} \frac{	an 	heta}{	heta} = 1$ નો ઉપયોગ કરતા: $= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$ $= 2 \cdot \frac{1}{4} \cdot (3 + 1)$ $= \frac{2}{4} \cdot 4 = 2$

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ ની કિંમત શોધો.

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