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Limit

lim x->+inf ln(1+xe^(2x))/sin^2(x) = ?
I expressed both the denominator and the nominator as taylor series...

lim x->0+ (ln(x))^5/x^(1/5) = ?
I tried evaluating it as lim x->0+ [x^(4/5) (ln(x))^5 / x] but it got me no where... I also tried expressing ln(x)^5 as e^5ln(lnx)... Doesn't work.

How should I approach these questions?

For the attached question...
How to prove b? By MI?? I couldn't prove that for n=k+1...

What is the application of having such "differential equation" = 0??

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Reply 1

ghostwalker

Original post by startanewww

Just looking at it, considering each term and what it evaluates to, what its range might be, can you see what the limit is going to be?.

Then

Spoiler

For the attached question...
How to prove b? By MI?? I couldn't prove that for n=k+1...

Yes, MI.
Post some working.

What is the application of having such "differential equation" = 0??

No idea.

(edited 9 years ago)

Reply 2

Protoxylic

Ignore this post, I was going to suggest L'Hospital's rule, but the limit is obviously not indeterminate.

(edited 9 years ago)

Reply 3

davros

Original post by startanewww

Is that first one supposed to be

\displaystyle \lim_{x \to \infty} \dfrac{ln(1 + xe^{2x})}{sin^2x}

?

Think about numerator and denominator separately as x gets larger and larger. Does it look like either or both tend to a limit? What does this say about the convergence or divergence of the overall expression?

Reply 4

startanewww

Original post by davros

Is that first one supposed to be

\displaystyle \lim_{x \to \infty} \dfrac{ln(1 + xe^{2x})}{sin^2x}

Thanks...
I just figured that e^(inf) and sin(inf) do not exist... Then I don't know what to do. I still haven't learnt convergence / divergence...

Reply 5

davros

Original post by startanewww

Thanks...
I just figured that e^(inf) and sin(inf) do not exist... Then I don't know what to do. I still haven't learnt convergence / divergence...

Not being rude, but if you haven't learned about convergence and divergence, why are you attempting this question?

The point with the first problem is: while e^x just gets bigger and bigger as x->infinity, sin x just oscillates between -1 and +1 with no fixed limit, and sin^2x oscillates between 0 and 1, so the overall expression doesn't tend to a limit.

Are you doing this for "fun" or as part of a course?

(edited 9 years ago)

Reply 6

james22

Original post by davros

I may be mistaken, but I think it does tend to a limit.

Reply 7

davros

Original post by james22

I may be mistaken, but I think it does tend to a limit.

Are you counting "positive infinity" as "tending to a limit"?

Reply 8

tombayes

Original post by startanewww

How to prove b? By MI?? I couldn't prove that for n=k+1...

b) is Leibniz rule - simple application

Original post by startanewww

What is the application of having such "differential equation" = 0??

Original post by ghostwalker

No idea.

allows you to find a power series solution at a point e.g. x=0 or perhaps more generally in some cases

(edited 9 years ago)

Reply 9

james22

Original post by davros

Are you counting "positive infinity" as "tending to a limit"?

Yes, although I just realised tat it isn't standard terminology.

Reply 10

davros

Original post by james22

Yes, although I just realised tat it isn't standard terminology.

That's OK.

I realized I was lecturing somebody on how to determine whether a limit existed or not when I hadn't checked what their definition of "converging" and "diverging" was, but the moral is always: find out what definition you're supposed to be using and stick to it :smile:

Reply 11

startanewww

Original post by davros

I'm studying a course called "University Mathematics". This question is in the revision exercise.

I just read about convergence and divergence. In this case, both the denominator and the nominator converges? I can't find what it converges to...

Reply 12

startanewww

Original post by tombayes

b) is Leibniz rule - simple application

allows you to find a power series solution at a point e.g. x=0 or perhaps more generally in some cases

Thank you for the info!!!!

Reply 13

davros

Original post by startanewww

You mean "numerator" I think :smile:

The numerator doesn't "converge" - it tends to +infinity because it just gets bigger and bigger as x does.

The denominator varies between 0 and 1 but always remains bounded by 1 as a maximum value.