Finding a limit through algebra

[chee]

Hi, I'm having a bit of a problem with a certain problem assigned to me...
I must find the limit of (lnx)/(x-1), as x approaches 1. However, I may not use L'Hopital's rule; I must stick to algebra methods to solve the problem, and apparently, the definition of limit is unnecessary.
Thanks for the help.




Note: as x approaches 0 has been changed to as x approaches 1

 

[napoleonmax]

This is actually easy if you forget everything about calc. for a sec. Looking at this equation simply, the numerator goes to negative infinity as x goes to zero and the denominator goes to a negative infitesimal number (-.00000000000...1). This fraction goes to positive infinity. Using l'Hôpital's Rule, f'(ln x) = 1/x and f'(x-1) = 1 therefore this limit is the same as 1/x/1= 1/x. I'm not sure if this is the ALGEBRAIC way to do this, but it certainly is a logical approach.

 

[robert Ihnot]

why not lift the whole to e^(Inx/(x-1) = x^(1/(x-1)), which is roughly 1/x for small x. Thus In( infinity) equals infinity is a good guess for the answer.

 

[stefanfuglsang]

Here is a hint:
Taylor series (from Schaum Mathematical Handbook):
ln(x) = 2[(x-1)/(x+1) + 1/3*(x-1)^2/(x+1)^2 + ...)

 

[HallsofIvy]

Hi, I'm having a bit of a problem with a certain problem assigned to me...
I must find the limit of (lnx)/(x-1), as x approaches 0. However, I may not use L'Hopital's rule; I must stick to algebra methods to solve the problem, and apparently, the definition of limit is unnecessary.
Thanks for the help.

L'Hopital's rule doesn't apply anyway- this is not of the form 0/0 or inf/inf, etc. As napoleonmax pointed out, it is of the form -inf/(-0) which means that the limit is +infinity (which is just a way of saying that it has no limit).

 

[chee]

type

The original limit problem was mistyped. Please note the change.

 

[HallsofIvy]

I'm not sure what you mean by "algebra methods". I would write ln(x) as a Taylor's series about x= 1.

 

[Galileo]

Here's another way:

Recall the definition of a derivative: f'(a)=lim (x->a) (f(x)-f(a))/(x-a)
You`ll see lim (x->1) ln(x)/(x-1) is the derivative of ln(x) at x=1.

 

[tomkeus]

\frac{lnx}{x-1}=lnx^\frac{1}{x-1} which is actually natural logarithm of x-1th root of x which converge to 1.

 

[turin]

ln(1⁰) = 1? I would think = 0.

I'm very curious to see (the meaning of) the algebraic method of taking a limit. Until today, I had always thought that a limit was essentially a calculus concept.

 

[tomkeus]

No it's lne because lnx^\frac{1}{x-1}=ln(1+x-1)^\frac{1}{x-1}
Make substitiution \frac{1}{x-1}=t. Then t is going towards infinity when x is going to 1. Expression becomes \lim_{t\rightarrow\infty} ln(1+1/t)^t which then converges to lne=1. If you don't believe me check with mathematica or something else.

 

[MathematicalPhysicist]

ln x=a
e^a=x
lim a/(e^a-1)
e^a->1
now because e^a approaches 1 a approaches 0
therefore we get 0/0.

i don't know if this the way you were searching for but at least it's algebraic approach.

 

[turin]

tomkeus,
OK, yes, that makes sense. I misread your previous post. It helps to see the latex.

LQG,
0/0 is not defined in algebra, is it?

 

[MathematicalPhysicist]

tomkeus,
OK, yes, that makes sense. I misread your previous post. It helps to see the latex.

LQG,
0/0 is not defined in algebra, is it?

that's my answer that it has no limit, doesn't converge.

 

[tomkeus]

You cannot solve this entirely by "algebra means". You can only reduce it to \lim_{x\rightarrow\infty} (1+1/x)^x. This is basic limit which cannot be proven with algebra only.