In the preface to William Dunham's book "Euler: The Master of Us All", he wrote,

 

[math]\lim_{t\rightarrow 0^{+}} \frac{1-x^t}{t} = -\ln x[/math] for [math]x > 0[/math]

 

Can anyone tell me how he got that result?

Edited November 8, 201015 yr by murshid

I will give you a hint.

 

Series expand [math]x^{t}[/math] about [math]t=0[/math] to first order in [math]t[/math] (you can go higher but that is all we will need). Put that expression into your limit and see what you get.

 

The only tricky thing here is working out what [math]\frac{\partial }{\partial t} x^{t}[/math] is. Either look it up, or see if you can prove it.

 

Let us know how you get on.

I will give you a hint.

 

Series expand [math]x^{t}[/math] about [math]t=0[/math] to first order in [math]t[/math] (you can go higher but that is all we will need). Put that expression into your limit and see what you get.

 

The only tricky thing here is working out what [math]\frac{\partial }{\partial t} x^{t}[/math] is. Either look it up, or see if you can prove it.

 

Let us know how you get on.

I managed to solve it using L'Hopital's rule.

 

[math]\frac{d}{dt} x^{t} = x^t \ln x[/math] (I got it by letting [math]y = x^t[/math], which is equivalent to [math]\ln y = t \ln x[/math], and then differentiating both sides with respect to t).

 

But what did you mean by Series expand [math]x^{t}[/math]? Did you mean the Taylor/Maclaurin Series expansion? I have been out of touch with calculus for the last few years. So it would really help if you could give me the series expansion of [math]x^{t}[/math] about [math]t = 0[/math].

Edited November 8, 201015 yr by murshid

I managed to solve it using L'Hopital's rule.

 

[math]\frac{d}{dt} x^{t} = x^t \ln x[/math]

 

But what did you mean by Series expand [math]x^{t}[/math]? Did you mean the Taylor/Maclaurin Series expansion? I have been out of touch with calculus for the last few years. So it would really help if you could give me the series expansion of [math]x^{t}[/math] about [math]t = 0[/math].

 

Right you have got it!

 

Now apply the Maclaurin series up to first order in [math]t[/math]. Near [math]t=0[/math] the function [math]x^{t}[/math] can be approximated by?

 

If you cannot remember the definition of the Taylor/Maclaurin series take a quick look at the Wikipedia entry.

Right you have got it!

 

Now apply the Maclaurin series up to first order in [math]t[/math]. Near [math]t=0[/math] the function [math]x^{t}[/math] can be approximated by?

 

If you cannot remember the definition of the Taylor/Maclaurin series take a quick look at the Wikipedia entry.

I think I've got it now. The expansion of [math]x^t[/math] near [math]t = 0[/math] is:

 

[math]1 + \frac{\ln x}{1}t + \frac{(\ln x)^2}{2}t^2 + \frac{(\ln x)^3}{6}t^3 + \cdots[/math]

 

Therefore,

 

[math] \lim_{t\rightarrow 0^{+}} \frac{1-x^t}{t} [/math]

 

[math]= \lim_{t\rightarrow 0^{+}} \frac{- \frac{\ln x}{1}t - \frac{\left (\ln x\right )^2}{2}t^2 - \frac{\left (\ln x\right )^3}{6}t^3 - \cdots}{t} [/math]

 

[math]= \lim_{t\rightarrow 0^{+}} - \frac{\ln x}{1} - \frac{\left (\ln x\right )^2}{2}t - \frac{\left (\ln x\right )^3}{6}t^2 - \cdots [/math]

 

[math]= - \ln x [/math]

Edited November 8, 201015 yr by murshid

You have it.

Can you tell me why the factorial sign isn't working in LaTex?

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Edited November 8, 201015 yr by murshid

I have had similar problems with LaTeX on here, minus signs have given me trouble. Try leaving more of a space between the number and the !.

I have had similar problems with LaTeX on here, minus signs have given me trouble. Try leaving more of a space between the number and the !.

I tried that. It doesn't work. For example, [math]\frac{\ln x}{1} [/math] works; but when I put a factorial sign after the '1' in the denominator, I get an error message: [math]\frac{\ln x}{1 !} [/math].

Edited November 8, 201015 yr by murshid

I have no idea, but the system does not like " ! ".

Edited November 8, 201015 yr by ajb

I have asked a question about the problem with the factorial sign here:

http://www.scienceforums.net/topic/3751-quick-latex-tutorial/page__view__findpost__p__572135

.

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I have a question about the original problem:

 

[math]\lim_{t\rightarrow 0^{+}} \frac{1-x^t}{t} = -\ln x[/math] for [math]x > 0[/math]

 

Why is it [math]t \rightarrow 0^{+}[/math] instead of [math]t \rightarrow 0^{-}[/math] or [math]t \rightarrow 0[/math]? I don't see why it shouldn't work for either [math]t \rightarrow 0^{-}[/math] or [math]t \rightarrow 0[/math].

Edited November 8, 201015 yr by murshid

The function looks smooth (fixing an x> 0). I think the direction of the limit won't matter.