"The value of $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $\le$ x is :"

Question

Accepted Answer

"$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}$$

$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}$$

$= {\cos ^{ - 1}}0 = {\pi \over 2}$"

The value of
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $\le$ x is :

Solution

About this question

Drill 25 more like these. Every day. Free.

The value of $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $\le$ x is :

Solution

About this question

More Limits, Continuity and Differentiability questions

Drill 25 more like these. Every day. Free.

The value of
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $\le$ x is :