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

$\lim _{n \rightarrow \infty} P\left(1+\frac{r}{100 n}\right)^{t n} =$

If $x_1 = 3$ and $x_{n+1} = \sqrt{2 + x_n}$ for $n \ge 1$,then $\lim_{n \to \infty} x_n$ is equal to

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {c^2}} - \sqrt {{x^2} + {d^2}} }} = $

The value of $\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module