JEE maths#10

Calculus Level 5

$\lim_{x\rightarrow\infty} \frac{2x^{1/2} + 3x^{1/3} +4x^{1/4}+...+nx^{1/n}}{(2x-3)^{1/2} + (2x-3)^{1/3} + ...+ (2x-3)^{1/n}}$

Find the value of the limit above to 3 decimal places.

For more JEE problems try my set 1
For KVPY practice questions try my set 2

The answer is 1.414.

3 solutions

Rahil Sehgal
Mar 30, 2017

Again Loved This One!!! ..

Md Zuhair - 4 years, 2 months ago

Thanks buddy :)

Rahil Sehgal - 4 years, 2 months ago

I wonder how we can evaluate this limit when $n$ approaches infinity

Sabhrant Sachan - 4 years, 2 months ago

Nuce question. will tgink about it

Md Zuhair - 4 years, 2 months ago

You should place the equation on top so that when the problem appears in the main page, which show a list of problems, the equation will appear instead of just "Find the value of...". Just the standard three dots instead 4 or 5. You can either key in "..." or "\cdots", c for centre. Leave a space between "my" and "set" and "set" and "1" and "2". "Upto" is not an English word. It should be "up to". In this case just "to" would do.

Chew-Seong Cheong - 4 years, 2 months ago

Okay sir I will take care of it next time.

Rahil Sehgal - 4 years, 2 months ago

Nice solution, got my vote.

Hana Wehbi - 4 years, 2 months ago

Aaghaz Mahajan
May 6, 2018

A JEE question deserves a JEE style answer.......simply put n = 2 and solve.......otherwise, refer to solution by @Rahil Sehgal

Giorgos K.
May 6, 2018

$Mathematica$

Limit[Sum[n*x^(1/n), {n,2,100}]/Sum[(2x-3)^(1/m),{m,2,100}],x->Infinity]

returns $\sqrt{2}$

JEE maths#10

The answer is 1.414.

3 solutions

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