# Math Help - Limit question

1. ## Limit question

I need help in calculating this limit:

lim x->e+ (logx)^(1/x-e)

This gives me an indeterminate form which is 1^inf. ( log(e)=1; 1/(e-e) = 1/0+ = inf )

I proceed to follow a process that always helped me in these indeterminate forms, which is applying the exponential:

lim x->e+ e^[log(logx)^(1/(1-e))] =

= lim x->e+ e^[(log(logx))/(x-e)]

Now, it's when I get confused:

I proceed to substitute the x for the e+, which gives me:

e^[(log(log(e))/(e-e)]

log(e) = 1; therefore log(log(e)) = log(1) = 0

e-e = 0; since it's a lateral limit form the right side, e-e=0+

Which gives me the result of e^(0/0+)

I don't know where to go from here. Is 0/0+ and indeterminate form? Did I do something wrong?

2. If everything is correct, then $\frac{0}{0.00000...1}=0$

$e^{0}=1$

3. Originally Posted by CMartins
I need help in calculating this limit:

lim x->e+ (logx)^(1/x-e)
let $\displaystyle y = \lim_{x \to e^+} (\log{x})^{\frac{1}{x-e}}$

$\displaystyle \log{y} = \lim_{x \to e^+} \frac{\log(\log{x})}{x-e}
$

L'Hopital ...

$\displaystyle \log{y} = \lim_{x \to e^+} \frac{1}{x\log{x}} = \frac{1}{e}$

$\displaystyle y = e^{\frac{1}{e}}$

4. I think, it should be something like this
lim x->e+ (logx)^(1/(x-e)) = lim x->e+ e^((1/(x-e))*log(log x)).
By L' Hospital's rule, this is
im x->e+ e^(1/(x*log(x))) = e^(1/e).

5. Originally Posted by CMartins
I need help in calculating this limit:

lim x->e+ (logx)^(1/x-e)

This gives me an indeterminate form which is 1^inf. ( log(e)=1; 1/(e-e) = 1/0+ = inf )

I proceed to follow a process that always helped me in these indeterminate forms, which is applying the exponential:

lim x->e+ e^[log(logx)^(1/(1-e))] =

= lim x->e+ e^[(log(logx))/(x-e)]

Now, it's when I get confused:

I proceed to substitute the x for the e+, which gives me:

e^[(log(log(e))/(e-e)]

log(e) = 1; therefore log(log(e)) = log(1) = 0

e-e = 0; since it's a lateral limit form the right side, e-e=0+

Which gives me the result of e^(0/0+)

I don't know where to go from here. Is 0/0+ and indeterminate form? Did I do something wrong?
Couldn't we do,

$y= (lnx)^{ \frac{1}{x-e}}$

$lny = \frac{1}{x-e} ln (lnx)$

$\lim_{x \to e^+ } lny = \frac { ln ( lnx ) }{x-e}$

This, of course is indeterminent, so apply Lhopitals to get

$\lim_{x \to e^+ } lny= \frac{ \frac{1}{lnx} \frac{1}{x} }{1}$

Which results in,

$\lim_{x \to e^+ } lny = \frac{1}{e}$

Since this is the limit of lny, the limit of y is simply,

$\lim_{x \to e^+ } y = e^{\frac{1}{e}}$

That seems to agree with wolfram, lim &#40;lnx&#41;&#94;&#40;1&#47;&#40;x-e&#41;&#41;, x&#61;e - Wolfram|Alpha

I am not very good with limits though! But I think this is good?

6. Ah! I get it now.
Thank you.