mathop {lim }limits_{n to infty } frac{{sqrt | vedclass

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

Question

(B) To evaluate the limit $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }}$,divide both the numerator and the den

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }}$,divide both the numerator and the denominator by $\sqrt{n}$:$\mathop {\lim }\limits_{n 	o \infty } \frac{\frac{\sqrt{n}}{\sqrt{n}}}{\frac{\sqrt{n}}{\sqrt{n}} + \frac{\sqrt{n+1}}{\sqrt{n}}} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{1 + \sqrt{1 + \frac{1}{n}}}$As $n 	o \infty$,the term $\frac{1}{n} 	o 0$.Therefore,the limit becomes $\frac{1}{1 + \sqrt{1 + 0}} = \frac{1}{1 + 1} = \frac{1}{2}$.

$\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }} = $

Similar Questions

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

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^n}}} \right)$ equals

$\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

If $n$ is an integer,then $\mathop {\lim }\limits_{x \to n + 0} (x - [x]) = $

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module