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

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

Question

(D) हमारे पास $\mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{n}{{n + y}}} ight)^n} = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{1

Vedclass · Accepted Answer

$e^{-y}$

Vedclass · Answer

(D) हमारे पास $\mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{n}{{n + y}}} ight)^n} = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{1}{1 + \frac{y}{n}}} ight)^n}$ है। $= \mathop {\lim }\limits_{n 	o \infty } {\left( {1 + \frac{y}{n}} ight)^{-n}}$ $= \mathop {\lim }\limits_{n 	o \infty } {\left[ {\left( {1 + \frac{y}{n}} ight)^n} ight]^{-1}}$ चूंकि $\mathop {\lim }\limits_{n 	o \infty } {\left( {1 + \frac{y}{n}} ight)^n} = e^y$,इसलिए: $= (e^y)^{-1} = e^{-y}$.

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{n}{{n + y}}} \right)^n}$ का मान ज्ञात कीजिए।

Similar Questions

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

$\mathop {\lim }\limits_{x \to \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right]$ का मान ज्ञात कीजिए।

सीमा ज्ञात कीजिए: $\mathop {\lim }\limits_{x \to 1} \frac{x^{2}+1}{x+100}$

$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)$ का मान है

$\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module