Lim-n - infty-1 - 2 - 3 - - n-n -2n-2 - is- 1-2-1-2-1-1-

The correct option is A 1/2 lim_n → infty[1 + 2 + 3 + ⋯ n/n +2n/2 ] = lim_n →∞[n (n +1)/(n +2)n/2 ] = lim_n →∞n/2[n+1 n 2/n + 2 ] = lim_n →∞n/2( 1/n +2 ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of Limit

lim_n → infty...

Question

${lim}_{n \to i n f t y} [\frac{1 + 2 + 3 + \dots n}{n + 2} \frac{n}{2}]$ is

$\frac{1}{2}$

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-1

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$- \frac{1}{2}$

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Solution

The correct option is D
$- \frac{1}{2}$

${lim}_{n \to i n f t y} [\frac{1 + 2 + 3 + \dots n}{n + 2} \frac{n}{2}]$
$= {lim}_{n \to \infty} [\frac{n (n + 1)}{(n + 2)} \frac{n}{2}] = {lim}_{n \to \infty} \frac{n}{2} [\frac{n + 1 - n - 2}{n + 2}]$
$= {lim}_{n \to \infty} \frac{n}{2} (\frac{- 1}{n + 2})$
$= {lim}_{n \to \infty} \frac{- 1}{2 (1 + \frac{2}{n})}$
$= \frac{- 1}{2 (1 + 0)} = - \frac{1}{2}$

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Similar questions

\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + (2 n - 1) - 2 n}{\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1}}

is equal to

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\frac{1}{2}

(b)

- \frac{1}{2}

Q. The value of

\lim_{n \to \infty} \{\frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2}\}

is

(a) 1/2

(b) 1

(c) −1

(d) −1/2

\lim_{n \to \infty} \frac{1^{2} + 2^{2} + 3^{2} + . . . + n^{2}}{n^{3}}

is equal to

(a) 1
(b) 1/2
(c) 1/3
(d) 0

Q. The value of

{lim}_{n \to \infty} (\frac{n}{(n + 1) \sqrt{2 n + 1}} + \frac{n}{(n + 2) \sqrt{2 (2 n + 2)}} + \frac{n}{(n + 3) \sqrt{3 (2 n + 3)}} . . . . . + \frac{1}{2 n \sqrt{3}})

Q. If

lim n \to \infty (\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + \frac{1}{3 + \sqrt{3 n}} + \dots + \frac{1}{2 n}) = {log}_{e} k

, then

k =

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limn→infty[1+2+3+⋯nn+2n2] is

The correct option is D −12 limn→infty[1+2+3+⋯nn+2n2] =limn→∞[n(n+1)(n+2)n2]=limn→∞n2[n+1−n−2n+2] =limn→∞n2(−1n+2) =limn→∞−12(1+2n) =−12(1+0)=−12

${lim}_{n \to i n f t y} [\frac{1 + 2 + 3 + \dots n}{n + 2} \frac{n}{2}]$ is