# Find the limit lim_{brightarrow1^-}int_0^bfrac{dx}{sqrt{1-x^2}}

Find the limit
$\underset{b\to {1}^{-}}{lim}{\int }_{0}^{b}\frac{dx}{\sqrt{1-{x}^{2}}}$
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Dora
Given:
$\underset{b\to {1}^{-}}{lim}{\int }_{0}^{b}\frac{dx}{\sqrt{1-{x}^{2}}}$
First, we will find integration:
$\underset{b\to {1}^{-}}{lim}{\int }_{0}^{b}\frac{dx}{\sqrt{1-{x}^{2}}}$
This is an standart integral:
$=\left[\mathrm{arcsin}\left(x\right){\right]}_{0}^{b}$
$=\left[\mathrm{arcsin}\left(b\right)-\mathrm{arcsin}\left(0\right)\right]$
$=\mathrm{arcsin}\left(b\right)$
Now, find the limit:
$\underset{b\to {1}^{-}}{lim}\mathrm{arcsin}\left(b\right)$
Plug b=1
$=\mathrm{arcsin}\left(1\right)$
$=\frac{\pi }{2}$
$\underset{b\to {1}^{-}}{lim}{\int }_{0}^{b}\frac{dx}{\sqrt{1-{x}^{2}}}=\frac{\pi }{2}$