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

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

Question

(B) ધારો કે $L = \lim _{x ightarrow 0}\left(\frac{1+	an x}{1+\sin x}ight)^{\operatorname{cosec} x}$.અહીં સ્વરૂપ $1^\infty$ હોવાથી,આપણે $\lim_{x \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) ધારો કે $L = \lim _{x ightarrow 0}\left(\frac{1+	an x}{1+\sin x}ight)^{\operatorname{cosec} x}$.અહીં સ્વરૂપ $1^\infty$ હોવાથી,આપણે $\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} (f(x)-1)g(x)}$ સૂત્રનો ઉપયોગ કરીશું.$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)}}$જ્યારે $x 	o 0$,ત્યારે $\cos x 	o 1$ અને $1-\cos x 	o 0$.$L = e^{\frac{0}{1(1+0)}} = e^0 = 1$.

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

Similar Questions

કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+x}-1}{x}$

$\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right)=$

કિંમત શોધો: $\mathop {\lim }\limits_{x \to 1} \frac{x^{15}-1}{x^{10}-1}$

$\lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right) = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module