# Sine inverse Limits ques

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Evaluate:

$\lim\limits_{x\to1} \frac{sin^{-1}(1- \sqrt x)}{1 -x}$

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$\lim\limits_{x\to1} \frac{sin^{-1}(1- \sqrt x)}{1 -x}$    $\left(\frac00- case\right)$

$= \lim\limits_{x\to1} \cfrac{\frac1{\sqrt{1 - (1 + x - 2\sqrt x)}}\times\frac{-1}{2\sqrt x}}{-1}$   (By using D.L.H. Rule)

$= \lim\limits_{x\to 1} \frac1{2\sqrt x \sqrt{2\sqrt x- x}}$

$= \frac1{2\sqrt{2-1}}$

$= \frac12$

by (11.1k points)
Please give a solution without applying the Rule
by (42.5k points)

$\lim\limits_{x\to 1}\frac{sin^{-1}(1-√x)}{1-x}$

= $\lim\limits_{ x \to 1}\cfrac{sin^{-1}(\frac{(1-\sqrt x)(1+\sqrt x)}{1+\sqrt x})}{1-x}$

= $\lim\limits_{ x \to 1}\cfrac{sin^{-1}(\frac{(1- x)}{1+\sqrt x})}{1-x}$

= $\lim\limits_{ x \to 1}\cfrac{(\frac{1- x}{1+\sqrt x})+\frac1{2\times3}(\frac{1-x}{1+√ x})^3 + \frac{1.3}{2\times4\times5}(\frac{1-x}{1+\sqrt x})^5+....}{1-x}$

(∵ sin-1x = x + 1/2 x3/3 + 1.3/2.4 x5/5 + ....)

= $\lim\limits_{ x \to 1}(\frac1{1+\sqrt x}+\frac1{6}\frac{(1-x)^2}{(1+\sqrt x)^3}+\frac3{40}\frac{(1-x^4)}{(1+\sqrt x)^5+...})$

= $\frac1{1+\sqrt1}+0$

= 1/2

by (11.1k points)
Is there any solution without using expansion of arcsinX ?