Problem with a complex limit

[avest]

Hello everyone! I have a problem with finding a limit as x approaches infinity [x -x²ln(1+1/x) ]
I' ve been trying to change x to lne^x then use the logharitms difference formula(dividing the arguments) then some transformations and de l'Hospital rule. I've been doing it for an hour and i just cannot solve it. Please, help me.

 

[Ishuda]

Hello everyone! I have a problem with finding a limit as x approaches infinity [x -x²ln(1+1/x) ]
I' ve been trying to change x to lne^x then use the logharitms difference formula(dividing the arguments) then some transformations and de l'Hospital rule. I've been doing it for an hour and i just cannot solve it. Please, help me.

What happens if you look at it as

[x -x²ln(1+1/x) ] = \(\displaystyle \frac{1 - x ln(1+\frac{1}{x})}{\frac{1}{x}}\)

and use l'Hospital's rule for 0/0 twice.

EDIT: Correct typo.

 

[pka]

What happens if you look at it as

[x -x²ln(1+1/x) ] = \(\displaystyle \frac{1 {\color{red}\large +} x ln(1+\frac{1}{x})}{\frac{1}{x}}\)

and use l'Hospital's rule for 0/0 twice.

There is a typo it should be minus to get 0/0.

 

[avest]

Thanks a lot!

 

[Steven G]

Hi,
Try factoring out an x and then you should she what the result is.
Did you try plugging in a very large number for x so that you have an idea what the limit should be?
Let us know what you get!
Jomo

 

[Ishuda]

Hi,
Try factoring out an x and then you should she what the result is.
Did you try plugging in a very large number for x so that you have an idea what the limit should be?
Let us know what you get!
Jomo

Just because I feel like fooling around: Consider
ln(1+t) = t - t²/2 + t³/3 - t⁴/4 + ...
and so
x² ln(1+1/x) = x - 1/2 + 1/(3x) - 1/(4x²) + ...
and
x-x²ln(1+1/x) = 1/2 - 1/(3x) [ 1 - 3/(4x) + 3/(5x²) - ...] = 1/2 - 1/(3x) f(x)
By the alternating series tests, f(x) converges and
1 - 3/(4x) < f(x) < 1
Thus
1/(3x) f(x) goes to zero as x goes to infinity and
\(\displaystyle \lim_{x\to\infty} x - x^2 ln(1+\frac{1}{x}) = \frac{1}{2}\)

 

[Steven G]

Just because I feel like fooling around: Consider
ln(1+t) = t - t²/2 + t³/3 - t⁴/4 + ...
and so
x² ln(1+1/x) = x - 1/2 + 1/(3x) - 1/(4x²) + ...
and
x-x²ln(1+1/x) = 1/2 - 1/(3x) [ 1 - 3/(4x) + 3/(5x²) - ...] = 1/2 - 1/(3x) f(x)
By the alternating series tests, f(x) converges and
1 - 3/(4x) < f(x) < 1
Thus
1/(3x) f(x) goes to zero as x goes to infinity and
\(\displaystyle \lim_{x\to\infty} x - x^2 ln(1+\frac{1}{x}) = \frac{1}{2}\)

Yes this works fine! I just do not see the point of factoring out f(x) as 1/(3x) and [ 1 - 3/(4x) + 3/(5x²) - ...] both go to 0 as x approaches oo. Hence their product approaches 0.
Good job!
Jomo

 

[Ishuda]

Yes this works fine! I just do not see the point of factoring out f(x) as 1/(3x) and [ 1 - 3/(4x) + 3/(5x²) - ...] both go to 0 as x approaches oo. Hence their product approaches 0.
Good job!
Jomo

Oh? And how would you prove that the limit is zero? By the same kind of alternating series test? But then I wouldn't have been fooling around as much which is what I was doing. Some like beef, some like pork, and some like nothing at all but I'm an anchovy man right now.

 

[Deleted member 4993]

Oh? And how would you prove that the limit is zero? By the same kind of alternating series test? But then I wouldn't have been fooling around as much which is what I was doing. Some like beef, some like pork, and some like nothing at all but I'm an anchovy man right now.

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