Limn-13-23-n3-n4-1-4l- Find l

lim_n→∞1^3+2^3+...+n^3/n^4=lim_n→∞n^4+n^2+2n^34n^4=14=14l So l=1 ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

limn→∞13+23+....

Question

${lim}_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{4}} = \frac{1}{4} l .$ Find $l$

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Solution

${lim}_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{4}} = l i m_{n \to \infty} \frac{n^{4} + n^{2} + 2 n^{3}}{4 n^{4}} = \frac{1}{4} = \frac{1}{4} l$

So $l = 1$

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Similar questions

lim n \to \infty [\frac{1^{3} + 2^{3} + 3^{3} + . . . . . . + n^{3}}{n^{4}}] =

Evaluate : ${lim}_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . . + n^{4}}{n^{5}} - {lim}_{n \to \infty} \frac{1^{3} + 2^{3} . . . . + n^{3}}{n^{5}}$

{lim}_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . + n^{4}}{n^{5}} - {lim}_{n \to \infty} \frac{1^{3} + 2^{3} + 3^{3} + . . . + n^{3}}{n^{5}}

is:

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

equals?

l i m_{n \to \infty} \frac{2^{3} - 1^{3}}{2^{3} + 1^{3}} . \frac{3^{3} - 1^{3}}{3^{3} + 1^{3}} . . . . \frac{n^{3} - 1}{n^{3} + 1}

equal

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limn→∞13+23+...+n3n4=14l. Find l

limn→∞13+23+...+n3n4=limn→∞n4+n2+2n34n4=14=14lSo l=1

${lim}_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{4}} = \frac{1}{4} l .$ Find $l$

${lim}_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{4}} = l i m_{n \to \infty} \frac{n^{4} + n^{2} + 2 n^{3}}{4 n^{4}} = \frac{1}{4} = \frac{1}{4} l$

So $l = 1$