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

(B) माना $A = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 2}}{{x + 1}}} ight)^{x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } {\le

Vedclass · Accepted Answer

$e$

Vedclass · Answer

(B) माना $A = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 2}}{{x + 1}}} ight)^{x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{1}{{x + 1}}} ight)^{x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } {\left[ {{{\left( {1 + \frac{1}{{x + 1}}} ight)}^{x + 1}}} ight]^{\frac{{x + 3}}{{x + 1}}}}$चूँकि $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{1}{{x + 1}}} ight)^{x + 1}} = e$ और $\mathop {\lim }\limits_{x 	o \infty } \frac{{x + 3}}{{x + 1}} = 1$,$A = e^1 = e$.

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ का मान है

Similar Questions

$\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(\tan^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ का मान है

$\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$ का मान ज्ञात कीजिए।

यदि $\lim _{n \rightarrow \infty} x^n \log _e x=0$ है,तो $\log _x 12=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module