1. ## Improper:

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

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.

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

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.

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