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

Question
Limits and continuity
Find the limit
$$\lim_{b\rightarrow1^-}\int_0^b\frac{dx}{\sqrt{1-x^2}}$$

2021-02-02
Given:
$$\lim_{b\rightarrow1^-}\int_0^b\frac{dx}{\sqrt{1-x^2}}$$
First, we will find integration:
$$\lim_{b\rightarrow1^-}\int_0^b\frac{dx}{\sqrt{1-x^2}}$$
This is an standart integral:
$$=[\arcsin(x)]_0^b$$
$$=[\arcsin(b)-\arcsin(0)]$$
$$=\arcsin(b)$$
Now, find the limit:
$$\lim_{b\rightarrow1^-}\arcsin(b)$$
Plug b=1
$$=\arcsin(1)$$
$$=\frac{\pi}{2}$$
$$\lim_{b\rightarrow1^-}\int_0^b\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}$$

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