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

[carlodelmundo]

Homework Statement

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

Homework Equations

L'Hopital's Rule
Algebraic Manipulation

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]

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.

To add to this, the limit could be infinity, -infinity, or not exist. If you have something like $$\lim_{x\to 0}\frac{1}{x}$$, 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]

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 :)

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.
$$\lim_{x \to 1^+}\frac{x}{ln(x)} = \infty$$
while
$$\lim_{x \to 1^-}\frac{x}{ln(x)} = -\infty$$

and yes it's legit :)

 

[Leptos]

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 :)

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

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

 

[icefirez]

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.
$$\lim_{x \to 1^+}\frac{x}{ln(x)} = \infty$$
while
$$\lim_{x \to 1^-}\frac{x}{ln(x)} = -\infty$$

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.