Help Solving: Euler's Limit Question (x > 0)

[murshid_islam]

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

$$\lim_{t\rightarrow 0^{+}} \frac{1-x^t}{t} = -\ln x$$ for $$x > 0$$

I can't figure out how the result comes out to be $-\ln x$. Can anyone help?

 

[arildno]

Use L'Hopital's rule

 

[murshid_islam]

Use L'Hopital's rule

Got it! Thanks!

I was trying the L'Hopital's rule before, but I was mistakenly trying it by differentiating with respect to x instead of t.

 

[arildno]

Got it! Thanks!

You're welcome!

I was trying the L'Hopital's rule before, but I was mistakenly trying it by differentiating with respect to x instead of t.

In general, in particular in non-linear optics, x-differentiation should never be chosen if a t-differentiation is possible.

 

[murshid_islam]

In general, in particular in non-linear optics, x-differentiation should never be chosen if a t-differentiation is possible.

Didn't get the joke.

 

[arildno]

It wasn't particularly funny, sort of an inductio-ad-aburdum, pythonesque sort of joke i often make, and that falls flat, SPLAT! on the ground...

 

[kahlan]

hello welcome
i think it is so easy, right
we will apply L'Hopital's rule for top and bottom

top = 1 - x^t ==> -x^t lnx
bottom = t ==> 1

then limit t----->0+ -x^t lnx / 1
t=0
so our limit will be -lnx
hi
-x^t = -x^0 =1

 

[mathwonk]

its just minus the derivative of x^t wrt t, at t=0, so the basic rule that the derivative of x^t is ln(x).x^t gives it, for t = 0.