Need help with a limit x->e (lnx - 1)/(x - e)

[johnk]

Hello,

I can't solve the following limit:

\(\displaystyle \L \lim_{x \to e}{ \frac{\ln x - 1}{x-e}}\)

(where ln is the natural logarithm)

The proffesor gave us a hint to make it into a limit that goes to zero (I think), but I wasn't able to do that.

 

[galactus]

I assume L'Hopital is not allowed?. That's a quick way out, but less interesting.

You could let \(\displaystyle \L\\u=ln(x)-1, \;\ e^{u+1}=x\)

\(\displaystyle \L\\\lim_{u\to\0}\frac{u}{e^{u+1}-e}\)

\(\displaystyle \L\\\lim_{u\to\0}\frac{u}{e^{u}e-e}\)

\(\displaystyle \L\\\lim_{u\to\0}\frac{u}{e(e^{u}-1)}\)

\(\displaystyle \L\\\frac{1}{e}\lim_{u\to\0}\underbrace{\frac{u}{e^{u}-1}}_{\text{limit is 1}}\)

\(\displaystyle \L\\=\fbox{\frac{1}{e}}\)

Now, to prove the last limit. It can be tricky(without L'Hopital). Taylor series is an option. But, a way I have used in the past is to express the derivative of e^x at x=0 as a limit. It's not a L'Hopital.

I will use h instead of u since that is the familiar notation.

\(\displaystyle \L\\\frac{d}{dx}[e^{x}]|_{x=0}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{1}{\frac{e^{0+h}-e^{0}}{h}}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{1}{\frac{e^{h}-1}{h}}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{h}{e^{h}-1}=\fbox{1}\)

 

[soroban]

Hello, John!

The other limit is: \(\displaystyle \L\:\lim_{x \to a}\,\frac{\sin x- \sin a}{x-a}\)

Here's my approach . . .

We need these two theorems:

. . . \(\displaystyle \L\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1\;\;\;\;\;\lim_{\theta\to0}\frac{1\,-\,\cos\theta}{\theta} \:=\:0\)


Let \(\displaystyle \,\theta\:=\:x\,-\,a\;\;\Rightarrow\;\;x\:=\:\theta\,+\,a\)

If \(\displaystyle x\to a\), then: \(\displaystyle \,\theta\to0\)

Then we have: \(\displaystyle \L\:\lim_{\theta\to0}\frac{\sin(\theta+a)\,-\,\sin a}{\theta}\)


We'll manipulate that fraction . . .

\(\displaystyle \L\frac{\sin(\theta+a)\,-\,\sin a}{\theta} \;=\;\frac{\sin\theta\cdot\cos a \,+\,\sin a\cdot\cos\theta \,-\,\sin a}{\theta}\)

. . \(\displaystyle \L= \;\frac{sin\theta\cdot\cos a \,-\, \sin a(1\,-\,\cos\theta)}{\theta} \;=\;\frac{\sin\theta\cdot\cos a}{\theta} \,-\,\frac{\sin a(1\,-\,\cos\theta)}{\theta}\)

. . \(\displaystyle =\;\left(\frac{\sin\theta}{\theta}\right)\cdot\cos a \,-\,\sin a\cdot\left(\frac{1\,-\,\cos\theta}{\theta}\right)\)


Then take the limit:

\(\displaystyle \L\lim_{\theta\to0}\left[\left(\frac{\sin\theta}{\theta}\right)\cdot\cos a \,-\,\sin a\cdot\left(\frac{1\,-\,\cos\theta}{\theta}\right) \right]\)

. . \(\displaystyle \L=\;\left(\lim_{\theta\to0}\,\frac{\sin\theta}{\theta}\right)\cdot \cos a \:-\:\sin a\cdot\left(\lim_{\theta\to0}\frac{1\,-\,\cos\theta}{\theta}\right)\)

. . \(\displaystyle \L= \;1\,\cdot\,\cos a \,-\,\sin a\,\cdot\,0\)

. . \(\displaystyle \L= \;\fbox{\cos a}\)

 

[johnk]

My second question was split into another topic (sorry for the trouble) and my "thank you" removed in the process, I'm not sure if Galactus saw it, so:
Thanks, Galactus!

Anyway, as I said, I hoped I'd be able to solve the other limit if I knew how to solve this one, but unfortunately not. So the other problem is here [solved now]: http://www.freemathhelp.com/forum/viewtopic.php?t=26508

And thank you, Soroban, for detailed help with the second limit, even though Stapel was faster and I figured it out with her help, using very similar approach to yours.

 

[pka]

Here is a purely algebraic way using properties of the logarithm. I suspect that what your instructor wanted. It is certainly what I would teach.
There is a well known inequality: \(\displaystyle \L 0 < a < b\quad \Rightarrow \quad \frac{1}{b} < \frac{{\ln (b) - \ln (a)}}{{b - a}} < \frac{1}{a}\)

Consider approaching e from the left:
\(\displaystyle \L 1 < x < e\quad \Rightarrow \quad \frac{1}{e} < \frac{{\ln (e) - \ln (x)}}{{e - x}} = \frac{{\ln (x) - 1}}{{x - e}}\)

Consider approaching from the right.
\(\displaystyle \L e < x\quad \Rightarrow \quad \frac{{\ln (x) - 1}}{{x - e}} < \frac{1}{e}\quad\).

So for x near e we get the limit of \(\displaystyle \L \frac{1}{e}\).

 

[johnk]

Here is a purely algebraic way using properties of the logarithm. I suspect that what your instructor wanted. It is certainly what I would teach.

This is interesting, but I don't think it's what he wanted. As I said, he hinted us to make it into a limit that goes to 0, which would fit Galactus' solution.

And to be honest, I don't think I've actually ever seen this inequality. Does it have a name? I'd have to know how to prove that first anyway, profs don't like us using theorems we can't prove.

 

[pka]

This is interesting, but I don't think it's what he wanted. As I said, he hinted us to make it into a limit that goes to 0, which would fit Galactus' solution.
I've actually ever seen this inequality. Does it have a name? I'd have to know how to prove that first anyway, profs don't like us using theorems we can't prove.

It is called Napier’s inequality. It is easily proved using the mean value theorem on \(\displaystyle f(x) = \ln (x)\). I have one problem with Galactus' solution. One still needs addition work to show that \(\displaystyle \lim _{u \to o} \frac{u}{{e^u - 1}} = 1\). If you are not using L'Hopital’s Rule then that is difficult.

 

[galactus]

I have one problem with Galactus' solution. One still needs addition work to show that \(\displaystyle \lim _{u \to o} \frac{u}{{e^u - 1}} = 1\). If you are not using L'Hopital’s Rule then that is difficult.

I thought someone may point that out. My bad for being lazy. Besides, I had to leave, then I hit a deer this morning and mashed my car up. So, I am back to finish. I emended my post to reflect a method I have used for these types.

 

[stapel]

I hit a deer this morning and mashed my car up.

Ouch! :shock:

Are you okay?

Eliz.

 

[galactus]

Thanks for asking, Stapel. Yes, I'm OK. But my little Toyota Yaris wasn't so lucky.

 

[pka]

I too am sorry to hear that. I lost three cars to deer. But I live in a county with 100 deer to ever person. We have on average 10 collisions between cars and deer every week!

 

[galactus]

I, too, live in an area like that(rural PA). Come to think of it, I've walloped a deer with just about every car I've had.

 

[Deleted member 4993]

I, too, live in an area like that(rural PA). Come to think of it, I've walloped a deer with just about every car I've had.

That could give rise somewhat "morbid" statistical problem .....

 

[morson]

If you are not using L'Hopital’s Rule then that is difficult.

What about this way?

\(\displaystyle \L\ \frac{u}{e^u - 1}\ = \frac{u}{(1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...) - 1} = \frac{u}{u + \frac{u^2}{2!} + \frac{u^3}{3!} + ...} = \frac{1}{1 + \frac{u}{2!} + \frac{u^2}{3!} + ...}\), \(\displaystyle \L\ u \not=\ 0\)

So:

\(\displaystyle \L\ \lim_{u \to 0} \frac{u}{e^u - 1}\ = \frac{1}{1 + \frac{0}{2!} + \frac{0^2}{3!} + ...} = \frac{1}{1}\ = 1\)

?