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

  • #1

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.
 

Answers and Replies

  • #2
Dick
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If you are getting the 'semi-indefinite form' 1/0, you can forget about manipulating it further. The limit doesn't exist.
 
  • #3
HallsofIvy
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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.
 
  • #4
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).
 
  • #5
2
0
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 :)
 
Last edited:
  • #6
gb7nash
Homework Helper
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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 [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.
 
Last edited:
  • #7
35,238
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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.
[tex]\lim_{x \to 1^+}\frac{x}{ln(x)} = \infty[/tex]
while
[tex]\lim_{x \to 1^-}\frac{x}{ln(x)} = -\infty[/tex]

and yes it's legit :)
 
  • #8
168
0
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 ex/lnx cannot be simplified unlike elnx = 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]
 
  • #9
2
0
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]


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.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
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966
There was nothing rude about Mark44's response.
 

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