Find Interesting Limit - Ideas Welcome!

[gomunkul51]

I found an interesting limit:

any ideas? :)

 

[Mark44]

This might be helpful.
$$\frac{(1 + 1/x)^{x^2}}{e^x} = \frac{(1 + 1/x)^{x^2}}{(e^{1/x})^{x^2}} = \left(\frac{1 + 1/x}{e^{1/x}}\right)^{x^2}$$

Let y = the last expression above, then take ln of both sides, then take the limit. Try to get something that you can use L'Hopital's Rule on.

I think I know what the limit is, but I haven't worked this all the way through.

 

[Gib Z]

Remember: $$\lim_{x\to\infty} \left( 1+ \frac{1}{x}\right)^x = e$$

 

[gomunkul51]

I can't seem to get to the right indeterminate form (inf/inf, 0/0) to use l'Hospital's rule.

Can someone please try to get the full result? :)

 

[Gib Z]

I can't seem to get to the right indeterminate form (inf/inf, 0/0) to use l'Hospital's rule.

Can someone please try to get the full result? :)

We can't give you answers like that.

You could try my suggestion.

EDIT: Perhaps my last hint was a bit too cryptic. Try taking the natural log of your limit and expand the log term into a series.

 

[njama]

$$y=\frac{(1+\frac{1}{x})^{x^2}}{e^x}$$

$$lny=x^2ln(1+\frac{1}{x})-x$$

Now take limit of both sides and write down the solution here.

Regards.

 

[gomunkul51]

njama: if I take the limit of both sides I get $$ln(y)=inf*0-inf$$ It doesn't make sense and you can't use l'Hopital's Rule on it.

Does anyone knows how to do it ?

 

[gomunkul51]

Here is a try:

$$exp( [ln(1 + 1/x) - 1/x] / [1/x^2] )$$

this is (I think) the correct type for l'Hopital's Rule of:
$$0/0$$

so I differentiate:
$$exp( [(x/[x+1]) + 1/x^2] / [-2/x^3] )$$

and this is basically:
$$exp(1/0) -> infinity$$

so limit should be infinity, but it's NOT.

I did it with changing the limit to (x goes to 0) and (x=1/x) here i could get the right answer.

I don't understand why can't I get the right limit in the first form !

 

[vela]

You didn't differentiate the log term correctly. It's easiest to do if you rewrite it as

$$\log \left(1+\frac{1}{x}\right) = \log \left(\frac{x+1}{x}\right) = \log (x+1) - \log x$$

and then differentiate with respect to x.

 

[gomunkul51]

My God, thank you very much vela !

It was driving me crazy!

probably the coming PDE test messed with my head a bit :)

 

[gomunkul51]

$$<br /> exp( [ln(1 + 1/x) - 1/x] / [1/x^2] )<br />$$

Differential:

$$<br /> exp( [(-1/[x^2+x]) + 1/x^2] / [-2/x^3] ) =<br /> = exp( -(1/2)*(x - x^2/x+1) ) = exp( -(1/2)*(x/x+1) ) = exp(-(1/2))<br /> <br /> =\frac{1}{e^1/2}$$

Which is the correct answer !

 

[Mark44]

Use njama's suggestion to write
$$lim~ln y = x^2 ln(1 + 1/x) - x = lim~x^2(ln(1 + 1/x) - 1/x) = lim~\frac{ln(1 + 1/x) - 1/x}{1/x^2}$$

(All limits taken as x --> infinity.)

As x --> infinity, the numerator --> 0 as does the denominator, so L' Hopital's Rule applies.
$$= lim~\frac{\frac{1}{(1 + 1/x)} \cdot (-1/x^2) + 1/x^2}{-2/x^3}$$

Evalulate the last limit and keep in mind that this is lim ln y, not lim y.

 

[Dickfore]

Take the log

$$\begin{array}{l}<br /> x^{2} \, \ln(1 + \frac{1}{x}) - x = x \, \left(x \ln(1 + \frac{1}{x}) - 1\right) \\<br /> <br /> = \frac{x \, \ln(1 + \frac{1}{x}) - 1}{\frac{1}{x}}<br /> \end{array}$$

Notice that:

$$\lim_{x \rightarrow \infty} \left[ x \ln(1 + \frac{1}{x})\right] \stackrel{t = 1\x}{=} \lim_{t \rightarrow 0} \frac{\ln(1 + t)}{t} = \frac{0}{0} = \lim_{t \rightarrow 0} \frac{1}{1 + t} = 1$$

so, you have an indeterminate form 0/0. Use L'Hospital's Rule.

Don't forget to exponentiate back for the final result.

 

[gomunkul51]

$$<br /> exp( [ln(1 + 1/x) - 1/x] / [1/x^2] )<br />$$

Differential:

$$<br /> exp( [(-1/[x^2+x]) + 1/x^2] / [-2/x^3] ) =<br /> = exp( -(1/2)*(x - x^2/x+1) ) = exp( -(1/2)*(x/x+1) ) = exp(-(1/2))<br /> <br /> =\frac{1}{e^1/2}$$



Which is the correct answer !

Thank you, but as you see I already did it !
Limit = exp(-(1/2))