# Finding a limit through algebra

1. Jun 24, 2004

### 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

Last edited: Jun 26, 2004
2. Jun 24, 2004

### 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.

3. Jun 24, 2004

### 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.

4. Jun 25, 2004

### 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 + ...)

5. Jun 25, 2004

### HallsofIvy

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).

6. Jun 26, 2004

### chee

type

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

7. Jun 26, 2004

### HallsofIvy

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

8. Jun 27, 2004

### 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.

9. Jul 5, 2004

### 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.

Last edited: Jul 6, 2004
10. Jul 5, 2004

### 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.

11. Jul 6, 2004

### 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 beleive me check with mathematica or something else.

Last edited: Jul 6, 2004
12. Jul 6, 2004

### 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 dont know if this the way you were searching for but at least it's algebraic approach.

13. Jul 6, 2004

### 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?

14. Jul 7, 2004

### MathematicalPhysicist

that's my answer that it has no limit, doesnt converge.

15. Jul 7, 2004

### 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.