Sine inverse Limits ques

Question

Evaluate:

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

Answer 1

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

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

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

\(= \frac1{2\sqrt{2-1}}\)

\(= \frac12\)

Comments

Please give a solution without applying the Rule

Answer 2

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

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

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

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

(∵ 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

Comments

Is there any solution without using expansion of arcsinX ?