lim _{n rightarrow infty} nleft(sqrt{n^2+9}-n | vedclass

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

Question

(D) To evaluate the limit $\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right)$,we rationalize the expression by multiplying and dividing by the

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(D) To evaluate the limit $\lim _{n ightarrow \infty} n\left(\sqrt{n^2+9}-night)$,we rationalize the expression by multiplying and dividing by the conjugate $\left(\sqrt{n^2+9}+night)$.$\lim _{n ightarrow \infty} n\left(\sqrt{n^2+9}-night) 	imes \frac{\sqrt{n^2+9}+n}{\sqrt{n^2+9}+n}$$= \lim _{n ightarrow \infty} \frac{n(n^2+9-n^2)}{\sqrt{n^2+9}+n}$$= \lim _{n ightarrow \infty} \frac{9n}{\sqrt{n^2+9}+n}$Divide the numerator and denominator by $n$:$= \lim _{n ightarrow \infty} \frac{9}{\sqrt{\frac{n^2+9}{n^2}}+1}$$= \lim _{n ightarrow \infty} \frac{9}{\sqrt{1+\frac{9}{n^2}}+1}$As $n ightarrow \infty$,$\frac{9}{n^2} ightarrow 0$,so the limit is $\frac{9}{\sqrt{1+0}+1} = \frac{9}{1+1} = \frac{9}{2}$.

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

Similar Questions

If $\mathop {\lim }\limits_{x \to 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$,where $n$ is a positive integer,then $n = $

Let $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$. The smallest positive integer $k$ such that $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is

If $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,and $S_3 = \sum_{r=1}^{n} r^3$,then the value of $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ is

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

$\mathop {\lim }\limits_{x \to \frac{\pi^+}{2}} e^{[\cot x]}$ is equal to :-
(where $[.]$ is the greatest integer function)

Vedclass Test Series

Exam Paper Generator

Online Exam Module