lim _{x rightarrow 0}left(frac{1+tan x}{1+sin | vedclass

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(B) माना $L = \lim _{x ightarrow 0}\left(\frac{1+	an x}{1+\sin x}ight)^{\operatorname{cosec} x}$.चूँकि यह $1^\infty$ का रूप है,हम सूत्र $\lim_{x

Vedclass · Accepted Answer

$1$

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(B) माना $L = \lim _{x ightarrow 0}\left(\frac{1+	an x}{1+\sin x}ight)^{\operatorname{cosec} x}$.चूँकि यह $1^\infty$ का रूप है,हम सूत्र $\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} (f(x)-1)g(x)}$ का उपयोग करेंगे।$L = e^{\lim _{x ightarrow 0}\left(\frac{1+	an x}{1+\sin x}-1ight) \operatorname{cosec} x}$$L = e^{\lim _{x ightarrow 0}\left(\frac{	an x-\sin x}{1+\sin x}ight) \cdot \frac{1}{\sin x}}$$L = e^{\lim _{x ightarrow 0}\left(\frac{\frac{\sin x}{\cos x}-\sin x}{1+\sin x}ight) \cdot \frac{1}{\sin x}}$$L = e^{\lim _{x ightarrow 0}\left(\frac{\sin x(1-\cos x)}{\cos x(1+\sin x)}ight) \cdot \frac{1}{\sin x}}$$L = e^{\lim _{x ightarrow 0}\frac{1-\cos x}{\cos x(1+\sin x)}}$जैसे $x 	o 0$,$\cos x 	o 1$ और $1-\cos x 	o 0$.$L = e^{\frac{0}{1(1+0)}} = e^0 = 1$.

$\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=$

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