Question

Prove that limit x tends to 0 sine inverse x upon x is equal to 1

Answer 1

${\underline{\sf{Question}}}$

Prove that $\sf\lim _{x \to0}\:\dfrac{\sin{}^{-1}x}{x}=1$

${\underline{\sf{Solution}}}$

$\sf\lim _{x \to0}\:\dfrac{\sin{}^{-1}x}{x}=1$

Since 0/0 is of indeterminate form.

Then ,apply L'Hospital's Rule.

$\sf\lim _{x \to0}\dfrac{\frac{d(\sin{}^{-1}x)}{dx}}{\frac{dx }{dx}}$

$\sf\lim _{x \to0}\dfrac{\frac{1}{\sqrt{1-x^2}}}{1}$

$\sf\lim _{x \to0}\dfrac{1}{\sqrt{1-x^2}}$

$\sf=\dfrac{1}{\sqrt{1-0}}$

$\sf=1$

Therefore,

$\sf\lim _{x \to0}\:\dfrac{\sin{}^{-1}x}{x}=1$

Hence Proved

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About L'HOSPITAL'S Rule :

If f(a)=g(a)= 0 , then

$\sf \lim _{x \to \: a} \: \frac{f(x)}{g(x)} = \sf \lim _{x \to \: a} \frac{f'(x)}{g'(x)}$