mathop {lim }limits_{x to infty } {left( {fra | vedclass

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

Question

(A) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 3}}{{x + 1}}} ight)^{x + 1}}$કૌંસની અંદરના પદને ફરીથી લખતા: $\frac{x+3}{x

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(A) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 3}}{{x + 1}}} ight)^{x + 1}}$કૌંસની અંદરના પદને ફરીથી લખતા: $\frac{x+3}{x+1} = \frac{x+1+2}{x+1} = 1 + \frac{2}{x+1}$હવે લક્ષ આ મુજબ થશે: $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{2}{x+1}} ight)^{x+1}}$પ્રમાણિત લક્ષના સૂત્ર $\mathop {\lim }\limits_{u 	o \infty } (1 + \frac{k}{u})^u = e^k$ નો ઉપયોગ કરતા,જ્યાં $u = x+1$ અને $k = 2$:$\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{2}{x+1}} ight)^{x+1}} = e^2$

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 3}}{{x + 1}}} \right)^{x + 1}} = $

Similar Questions

$\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{(\operatorname{cosec} x-\cot x)(e^x-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}} = $

લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 3} (x+3)$

$\mathop {\lim }\limits_{x \to \pi /2} \frac{{1 + \cos 2x}}{{{{(\pi - 2x)}^2}}} = $

$\mathop {\lim }\limits_{x \to 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module