Limits again

JEE Mains 2015 Guidance › Calculus questions › Limits again

Lt nâ†’âˆž Lt xâ†’0 1/(1^tan²x+2^tan²x+....N^tan²x)^cot²x

We had a similar problem
http://www.targetiit.com/iit-jee-forum/posts/find-the-limit-10114.html

But I'm not sure about the answer...Plz someone give a complete explanation with answer :)

#Mathematics #Calculus

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2 Answers

kartik sondhi ·2009-08-13 09:39:24

as Lt xâ†’0 by keeping11^tan²x+2^tan²x1^tan³x..âˆž=y then Take log on both sides
log 1y= cot²xlog(tan²x+2^tan²x1^tan³x..âˆž)
then Take cot²x as 1tan²x
Now Diffrentiate on both sides As L'Hospital rule

and use the tan Expansion formula

then You will get the answer

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Devil ·2009-08-13 09:43:16

Well, I think limit is 1.We will probably have lnk=cot²xln1, meaning k=1...

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Basic Integral Calculus Operators Symbols Matrix Cases

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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Limits again

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