lim _{n rightarrow infty}left{n-sqrt{n^2-4 n} | vedclass

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(B) To evaluate the limit $\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}$,we multiply and divide by the conjugate expression:$= \lim _{n

Vedclass · Accepted Answer

$2$

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(B) To evaluate the limit $\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}$,we multiply and divide by the conjugate expression:$= \lim _{n \rightarrow \infty} \frac{(n-\sqrt{n^2-4 n})(n+\sqrt{n^2-4 n})}{n+\sqrt{n^2-4 n}}$$= \lim _{n \rightarrow \infty} \frac{n^2-(n^2-4 n)}{n+\sqrt{n^2-4 n}}$$= \lim _{n \rightarrow \infty} \frac{4 n}{n+\sqrt{n^2-4 n}}$Divide the numerator and denominator by $n$:$= \lim _{n \rightarrow \infty} \frac{4}{1+\sqrt{1-\frac{4}{n}}}$As $n \rightarrow \infty$,$\frac{4}{n} \rightarrow 0$,so the expression becomes:$= \frac{4}{1+\sqrt{1-0}} = \frac{4}{1+1} = \frac{4}{2} = 2$.

$\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}=$

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