Improper:

**^_^Engineer_Adam^_^** · September 12th 2007, 08:49 AM

My hardest and last problem about I.I. :

0 has a discontinuity here so:
$\int_{0}^{1} \frac{dx}{x\sqrt{4-x^2}}$
$\lim_{t\to 0} \int_{t}^{1} \frac{dx}{x\sqrt{4-x^2}}$

to make it short, i used trigo substitution:
and im having trouble with this part

$\lim_{t \to 0} \frac{1}{2} \ln(\frac{2}{x} - \frac{\sqrt{4-x^2}}{x}) \big|_{t}^{1}$

$\lim_{t \to 0} \frac{1}{2} (\ln(2-\sqrt{4-x}) - \ln(x)) \big|_{t}^{1}$

$=\frac{1}{2} \ln(2-\sqrt{4-1}) - \frac{1}{2}\ln(1) - \frac{1}{2}\ln(2-2) + \frac{1}{2}\ln(0)$

which is
$\frac{1}{2} \ln(2-\sqrt{4-1}) - \frac{1}{2}\ln(1) - \infty + \infty$

i am having trouble because its indeterminate:
but i do not know how LHR works with ln

help
and thank you so much for the help on the other one

**Jhevon** · September 12th 2007, 09:40 AM

Originally Posted by ^_^Engineer_Adam^_^

My hardest and last problem about I.I. :

0 has a discontinuity here so:
$\int_{0}^{1} \frac{dx}{x\sqrt{4-x^2}}$
$\lim_{t\to 0} \int_{t}^{1} \frac{dx}{x\sqrt{4-x^2}}$

to make it short, i used trigo substitution:
and im having trouble with this part

$\lim_{t \to 0} \frac{1}{2} \ln(\frac{2}{x} - \frac{\sqrt{4-x^2}}{x}) \big|_{t}^{1}$

$\lim_{t \to 0} \frac{1}{2} (\ln(2-\sqrt{4-x}) - \ln(x)) \big|_{t}^{1}$

$=\frac{1}{2} \ln(2-\sqrt{4-1}) - \frac{1}{2}\ln(1) - \frac{1}{2}\ln(2-2) + \frac{1}{2}\ln(0)$

which is
$\frac{1}{2} \ln(2-\sqrt{4-1}) - \frac{1}{2}\ln(1) - \infty + \infty$

i am having trouble because its indeterminate:
but i do not know how LHR works with ln

help
and thank you so much for the help on the other one

i don't believe it is indeterminate, the limit just diverges.

if you leave it as $\lim_{t \to 0} \frac {1}{2} \left[ \ln \left| \frac {2 - \sqrt {4 - x^2}}{x} \right| \right]_{t}^{1}$

we have that $\lim_{t \to 0} \frac {2 - \sqrt {4 - t^2}}{t} = 0$

so the last term goes to ln0, which diverges.

**^_^Engineer_Adam^_^** · September 12th 2007, 10:00 AM

but the last term is $\ln \frac{2-\sqrt{4-0}}{0}$
$= \ln{\frac {0}{0}}$
or am i forgetting the rules of lims

**Jhevon** · September 12th 2007, 10:04 AM

Originally Posted by ^_^Engineer_Adam^_^

but the last term is $\ln \frac{2-\sqrt{4-0}}{0}$
$= \ln{\frac {0}{0}}$

Recall that ln is continuous on its domain.

so $\lim_{t \to 0} \frac {1}{2} \ln \left| \frac {2 - \sqrt {4 - t^2}}{t} \right| = \frac {1}{2} \ln \left| \lim_{t \to 0} \frac {2 - \sqrt {4 - t^2}}{t} \right| = \frac {1}{2} \ln |0| \rightarrow - \infty$ , when we subtract this, we add $\infty$ , so we diverge (we can find that limit by L'Hopital's rule... though everyone around here hates it.

**TKHunny** · September 12th 2007, 01:10 PM

Originally Posted by ^_^Engineer_Adam^_^

but the last term is $\ln \frac{2-\sqrt{4-0}}{0}$
$= \ln{\frac {0}{0}}$
or am i forgetting the rules of lims

As you appear to be simply and casually stuffing in the value that you shouldn't be stuffing in, there would be no other rational conclusion other than as you have stated it. You most certainly are forgetting the rules of limits.

Rule #1: The limiting value is NOT included in the limiting process. It might not even be in the Domain! Substitution is inappropriate. Don't do it.

Rule #2: If you later find the function to be continuous at the limiting value, ignore Rule #1 and go ahead with the substitution.

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