Limit calculation problem

A

akleron

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Dec 28, 2019
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Hello !
I'm trying to figure out this limit for a while but can't see a way to solve it.
I know that the limit of (1+1/x)^x as x->inf is equal to e so I got " infinity times zero" kind of limit.
It sounds like l'hopital and we did recently learn that law, But I only know how to use it when i have limit in the form of fractions.
I did try to rewrite the limit of f(x) as limit of e^ln[f(x)] but I didnt get too far..

Any idea what should I do here ?

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Try rewriting the expression as [MATH]\frac{(1 + \frac{1}{x})^x - e}{\frac{1}{x}}[/MATH]. Then apply l'Hopital.
 
Dr.Peterson said:
Try rewriting the expression as [MATH]\frac{(1 + \frac{1}{x})^x - e}{\frac{1}{x}}[/MATH]. Then apply l'Hopital.
Click to expand...

This is what I managed to do so far, should I go for another l'hopital ? can't see how its gonna turn solvable .. :/
1579458461949.png
 
akleron said:
How can I benefit from it ?
Can u explain please ?
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In my opinion, whenever faced with a "complicated" limit problem - graphing should be considered, specially when it is easy and free through WA (pka's favorite).
 
akleron said:
This is what I managed to do so far, should I go for another l'hopital ? can't see how its gonna turn solvable .. :/
View attachment 16173
Click to expand...
Your derivative in the numerator is slightly wrong; that shouldn't be [MATH]e^x[/MATH]. But that doesn't make it any simpler.

I haven't worked it out yet; but the next thing I would do would be to rearrange it in some way that might make another application of l'Hopital profitable. Another thing to consider (actually my thought when I made my first suggestion) is to substitute y = 1/x before applying l'Hopital at all. That's what I'm about to pursue ...

Before I work on that, though, we should ask where this came from. It is indeed complicated; so we wonder whether it is from a class and meant to teach you something (such as perseverance!), or has some use (so that just getting the answer would be enough), or is a sample contest problem for which technology is not available and you have to prove your answer.

ADDITION:
In the process, you may find it useful to use the known fact that [MATH](1 + \frac{1}{x})^x[/MATH], or [MATH](1 + y)^\frac{1}{y}[/MATH], approaches [MATH]e[/MATH], and also the fact that [MATH]\frac{\ln(1+y)}{y}[/MATH] approaches 1.
 
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Dr.Peterson said:
Your derivative in the numerator is slightly wrong; that shouldn't be [MATH]e^x[/MATH]. But that doesn't make it any simpler.

I haven't worked it out yet; but the next thing I would do would be to rearrange it in some way that might make another application of l'Hopital profitable. Another thing to consider (actually my thought when I made my first suggestion) is to substitute y = 1/x before applying l'Hopital at all. That's what I'm about to pursue ...

Before I work on that, though, we should ask where this came from. It is indeed complicated; so we wonder whether it is from a class and meant to teach you something (such as perseverance!), or has some use (so that just getting the answer would be enough), or is a sample contest problem for which technology is not available and you have to prove your answer.

ADDITION:
In the process, you may find it useful to use the known fact that [MATH](1 + \frac{1}{x})^x[/MATH], or [MATH](1 + y)^\frac{1}{y}[/MATH], approaches [MATH]e[/MATH], and also the fact that [MATH]\frac{\ln(1+y)}{y}[/MATH] approaches 1.
Click to expand...

Hello
Thanks for all the efforts !
Well, this question is from a task i got this week
we are not allowed to use technology and we must prove the answer .
 
Keep at it, then.

You can guess that I did succeed, and my added hints were ideas I used (though you may find a different way); let's see your next attempt.
 
Before you use l'Hopital you must note that \(\displaystyle (1 + \frac{1}{x})^x\) tends to e as x goes to \(\displaystyle \infty\), else you do not see \(\displaystyle \infty\)*0 !
 
Dr.Peterson said:
Keep at it, then.

You can guess that I did succeed, and my added hints were ideas I used (though you may find a different way); let's see your next attempt.
Click to expand...

I had many attempts using your tips, they did help and I think I got close to the solution.
But couldn't finish it yet.
Any further help ?
1579526478908.png
 
What you have agrees exactly with a middle stage of my work, except that you dropped a negative sign from the previous line.

Now observe that \(\displaystyle e^{\frac{\ln(1+t)}{t}} \)approaches e, by either of the facts I mentioned in post #8, so if the other factor has a limit, the limit you want will be that times e.

Apply l'Hopital to that first factor (after, of course, verifying that it applies). I'd expand the denominator before differentiating it.
 
Dr.Peterson said:
What you have agrees exactly with a middle stage of my work, except that you dropped a negative sign from the previous line.

Now observe that \(\displaystyle e^{\frac{\ln(1+t)}{t}} \)approaches e, by either of the facts I mentioned in post #8, so if the other factor has a limit, the limit you want will be that times e.

Apply l'Hopital to that first factor (after, of course, verifying that it applies). I'd expand the denominator before differentiating it.
Click to expand...

I think I got it now..
Thank you !!

One question though, how did you know that keeping on with l'hopital is the way ? are there any signs suggesting that ?

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I think there's a little sign error in the middle that turned out not to matter. Your answer is correct.

How did I know to continue with l'Hopital? Mostly because I could see nothing else to try! But I also kept my eye open for other things to do (like that split into a product, and something else I'll mention in a moment), in order to simplify things before I had to differentiate something ugly.

I didn't use l'Hopital the last time; that's where I recognized that the limit of ln(t+1)/t is 1, so I could factor that out (I didn't have the 2 there, which is a mistake) and be left with 1/(3t + 2), which approaches 1/2. What you did is fine at that point, and no harder than mine.
 
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