Question

${lim}_{x\to 0}\frac{\sqrt{1-cos2x}}{\sqrt{2x}}is$

• A. λ
• B. -1
• C. zero
• D. Does not exist

Answer 1

The correct option is D

Does not exist

$\underset{x\to 0}{lim}\frac{\sqrt{1-cos2x}}{\sqrt{2x}}=\underset{x\to 0}{lim}\frac{\sqrt{1-(1-2si{n}^{2}x)}}{\sqrt{2x}}\underset{x\to 0}{lim}\frac{\sqrt{2si{n}^{2}x}}{\sqrt{2x}}\underset{x\to 0}{lim}\frac{|sinx|}{x}\text{Let f}\left(x\right)=\frac{|sinx|}{x}\text{then, f}\left(0^{+}\right)=\underset{x\to 0}{lim}\frac{|sin(0+h)|}{h}=\underset{x\to 0}{lim}\frac{sinh}{h}=1\text{and}f\left(0^{-}\right)=\underset{x\to 0}{lim}\frac{|sin(0-h)|}{-h}=\underset{x\to 0}{lim}\frac{sinh}{-h}=-1\therefore \text{f}\left(0^{+}\right)\ne \text{f}\left(0^{-}1\right)\therefore \text{The limit does not exist}$