The correct option is D We have lim_x→∞[∑ n^2/n^3] We know, ∑ n^2=n(n+1)(2n+1)/6 lim_x→∞n(n+1)(2n+1)/6n^3 lim_x→∞1/6(n/n) (n+1/n) (2n+1/n) lim_x→∞1/6(1+1/n ...

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Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

lim x →∞[∑ n2...

Question

$lim x \to \infty [\frac{\sum n^{2}}{n^{3}}]$

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Solution

The correct option is C

We have $lim x \to \infty [\frac{\sum n^{2}}{n^{3}}]$

We know, $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

$lim x \to \infty \frac{n (n + 1) (2 n + 1)}{6 n^{3}}$

$lim x \to \infty \frac{1}{6} (\frac{n}{n}) (\frac{n + 1}{n}) (\frac{2 n + 1}{n})$

$lim x \to \infty \frac{1}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n})$

$= \frac{1}{6} (1 + 0) (2 + 0)$

$= \frac{1}{6} \times 2 = \frac{1}{3}$

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Similar questions

l i m x \to \infty (\frac{1}{n^{3} + 1} + \frac{4}{n^{3} + 1} + \frac{9}{n^{3} + 1} + \dots \dots + \frac{n^{2}}{n^{3} + 1})

is equal to

Q. Evaluate the limit to infinity

lim x \to \infty \frac{1^{2} + 2^{2} + - - - - + n^{2}}{n^{3}} = \frac{1}{3}

Q. If

α = lim n \to \infty (\frac{1}{n^{3} + 1} + \frac{4}{n^{3} + 1} + \frac{9}{n^{3} + 1} + \dots + \frac{n^{2}}{n^{3} + 1})

and

β = lim x \to 0 \frac{sin 2 x}{sin 8 x},

then a quadratic equation whose roots are

α

and

β

Evaluate:

lim n \to \infty [\frac{1}{n^{3}} + \frac{2^{2}}{n^{3}} + \frac{3^{2}}{n^{3}} + \dots + \frac{n^{2}}{n^{3}}]

Q. If

C_{1}, C_{2}, C_{3}

.....represent the speeds of

n_{1}, n_{2}, n_{3}

.....molecules, then the root mean square speed is
(a)

{(\frac{n_{1} C_{1}^{2} + n_{2} C_{2}^{2} + n_{3} C_{1}^{3} + \dots}{n_{1} + n_{2} + n_{3} + \dots})}^{\frac{1}{2}}

(b)

\frac{(n_{1} C_{1}^{2} + n_{2} C_{2}^{2} + n_{3} C_{1}^{3} + \dots)^{\frac{1}{2}}}{n_{1} + n_{2} + n_{3} + \dots}

(c)

\frac{(n_{1} C_{1}^{2})^{\frac{1}{2}}}{n_{1}} + \frac{(n_{2} C_{2}^{2})^{\frac{1}{2}}}{n_{2}} + \frac{(n_{3} C_{1}^{3})^{\frac{1}{2}}}{n_{3}} + \dots

(d)

{[\frac{(n_{1} C_{1}^{2} + n_{2} C_{2}^{2} + n_{3} C_{1}^{3} + \dots]^{2}}{(n_{1} + n_{2} + n_{3} + \dots)}]}^{\frac{1}{2}}

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limx→∞[∑n2n3]

The correct option is C We have limx→∞[∑n2n3] We know, ∑n2=n(n+1)(2n+1)6 limx→∞n(n+1)(2n+1)6n3 limx→∞16(nn)(n+1n)(2n+1n) limx→∞16(1+1n)(2+1n) =16(1+0)(2+0) =16×2=13

$lim x \to \infty [\frac{\sum n^{2}}{n^{3}}]$