The limit as x approaches 1 of x / ln (x)

carlodelmundo

1. The problem statement, all variables and given/known data

Hi. I'm having problems with the limit of x --> 1 of ( x / ln(x) ).

2. Relevant equations

L'Hopital's Rule
Algebraic Manipulation

3. The attempt at a solution

I understand that the the limit will give me a semi-indeterminate form (that is, it's answer is 1 / 0).

What I don't understand is how I can manipulate ln (x) so it's not in the denominator. I tried thinking of ways to use e^x.... but realized that multiplying by e^x does nothing. I'm presuming that we can raise both the numerator and the denominator by e^x but was not sure if it was legitimate.

Thanks in advance.

Dick

If you are getting the 'semi-indefinite form' 1/0, you can forget about manipulating it further. The limit doesn't exist.

HallsofIvy

Any time you have a ratio of two continuous functions, and just putting the value gives 1/0 (or more generally any non-zero value over 0) the limit does not exist. As x gets closer and closer to the "target value" the numerator stays close to 1 while the denominator gets close to 0. I.e. the fraction gets larger and larger. There is no limit.

carlodelmundo

Thanks, both. I understand it now. I made a numerical table to estimate the limit.... and as you two pointed out... the limit does not exist (infinite discontinuities).

icefirez

well i don't think so maybe I'm wrong but let's see

lim as x->1 (x/lnx) now me remove the natural lag

lim as x->1 ( e^x/ x) so as X approaches 1 we get lim x->1 (e^1/1)=e :)

and yes it's legit :)

gb7nash

To add to this, the limit could be infinity, -infinity, or not exist. If you have something like [tex]\lim_{x\to 0}\frac{1}{x}[/tex], x could approach 0 from the left, from the right, or from both directions. In this case, we can't put a definitive answer.

Note: This only works if you include the extended real line. If not, then it wouldn't exist at all.

Mark44

What makes you think that you can remove the natural "lag" (sic) in this way?

As was already said, this limit does not exist. The two one-sided limits are as far apart as they could possibly be.
[tex]\lim_{x \to 1^+}\frac{x}{ln(x)} = \infty[/tex]
while
[tex]\lim_{x \to 1^-}\frac{x}{ln(x)} = -\infty[/tex]

Leptos

Actually e^x/lnx cannot be simplified unlike e^lnx = x. For your second expression you'll get:

[tex]
\lim_{x \to 1^+}e^(x/lnx) = \infty
[/tex]
[tex]
\lim_{x \to 1^-}e^(x/lnx) = 0
[/tex]

icefirez

sorry yes I'm wrong... but you don't have to be rude and I know that you have to write log instead of "LAG" but please it's not the end of world.

HallsofIvy

There was nothing rude about Mark44's response.