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

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(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

$\frac{9}{2}$

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(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 $\frac{9}{\sqrt{1+0}+1} = \frac{9}{1+1} = \frac{9}{2}$.

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

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