LimitWatch

Announcements
#1
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??
0
4 years ago
#2
(Original post by startanewww)
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?
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:
Show

find a lower bound for the formula.

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.
0
4 years ago
#3
Ignore this post, I was going to suggest L'Hospital's rule, but the limit is obviously not indeterminate.
0
4 years ago
#4
(Original post by startanewww)
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??
Is that first one supposed to be

?

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?
0
#5
(Original post by davros)
Is that first one supposed to be

?

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?
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...
0
4 years ago
#6
(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?
0
4 years ago
#7
(Original post by davros)
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?
I may be mistaken, but I think it does tend to a limit.
0
4 years ago
#8
(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"?
0
4 years ago
#9
(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
2
4 years ago
#10
(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.
0
4 years ago
#11
(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
0
#12
(Original post by davros)
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?
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...
0
#13
(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!!!!
0
4 years ago
#14
(Original post by startanewww)
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...
You mean "numerator" I think

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.
0
#15
Then can I conclude that the limit does not exist, since the numerator tends to infinity and/or denominator = 0?
0
4 years ago
#16
(Original post by startanewww)
Then can I conclude that the limit does not exist, since the numerator tends to infinity and/or denominator = 0?
Correct, although personally I would say that "the expression tends to infinity as x tends to infinity"
1
X

new posts
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• Solent University
Mon, 20 May '19
• London Metropolitan University
Tue, 21 May '19
• Brunel University London
Wed, 22 May '19

Poll

Join the discussion

How has 2019 been so far?

Amazing!!! (39)
5.74%
Fairly positive (226)
33.28%
Just another year... (268)
39.47%
Is it 2020 yet? (146)
21.5%