Question

The following inequality is true for all x close to 0.

2-$\frac{x^2}{3}$ < $\frac{xsinx}{1-cosx}<2$

What is the value of ${lim}_{x\to 0}\frac{xsinx}{1-cosx}$?

• A. 2
• B. 1/2
• C. 0
• D. 1

Answer 1

The correct option is A 2

${lim}_{x\to 0}\frac{xsinx}{1-cosx}$

It is $\frac{0}{0}$ form

L's hospital rule (differentiate)

$\Rightarrow {lim}_{x\to 0}\frac{xcosx+sinx}{-(-sinx)}$

Putting 0 everywhere $\Rightarrow$ still $\frac{0}{0}$ form.

Differentiating numerator and denominator again,

${lim}_{x\to 0}\frac{x(-sinx)+cosx+cosx}{cosx}=2$