# Where have i gone wrong in solving this limit problem?

Problem: Evaluate, if the limit exists: $$\lim_{x\rightarrow 0}\frac{\sqrt{\frac{1}{2}(1-\cos2x)}}{x}$$

My attempt:

$\lim_{x\rightarrow 0}\frac{\sqrt{\frac{1}{2}(1-\cos2x)}}{x}$

$=\lim_{x\rightarrow 0}\frac{\sqrt{\frac{1}{2}(2\sin^2 x)}}{x}$

$=\lim_{x\rightarrow 0}\frac{\sqrt{\sin^2 x}}{x}$

$=\lim_{x\rightarrow 0}\frac{\sin\ x}{x}$

Since there is a standard formula for $\lim_{x\rightarrow 0}\frac{\sin\ x}{x}$, this limit exists and is equal to 1.

My problem: The answer given in my book is that the limit does not exist. There is a hint at the end of the problem which asks to use the concept of absolute value. I cannot understand why my method is not correct and by the way there is no absolute value in the expression.

• $$\sqrt{u^2} = |u|$$ Jul 8 '17 at 21:19
• Those limits does not $\implies$ each other, to the extent your calculations are correct, they are equal. Jul 8 '17 at 21:19
• @quasi, how is $\sqrt{u^2} = |u|$?
– MrAP
Jul 8 '17 at 21:20
• You are right @Henrik. Thanks for pointing out. Corrected.
– MrAP
Jul 8 '17 at 21:22
• @MrAP: Let u = -3. Then for that value of $u$ $$\sqrt{u^2} = \sqrt{9} = 3 = -u$$ Jul 8 '17 at 21:24

Hint: Note that $\sqrt{\sin^2 x} = |\sin x|$.

By definition
$$|x|=\begin{cases}x,&x\ge0 \\ -x,& x<0.\end{cases}$$ With this definition it is straightforward to show that $|x| = \sqrt{x^2}$ for any real number $x$:

Suppose $n\ge0$. By definition $\sqrt n$ is the principal root of $n$. That is, $\sqrt n\ge0$ and $(\sqrt n)^2=n$. Thus, if $x\ge0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $-x>0$ and so $\sqrt{x^2} = -x = |x|$.

• Why is $\sqrt{\sin^2 x} = |\sin x|$?
– MrAP
Jul 8 '17 at 21:19
• Because the square root of a positive number always is positive. Jul 8 '17 at 21:21
• Because the sqrt symbol delivers only the principal square root (the nonnegative one), hence when $\sin(x)$ is negative $$\sqrt{\sin^2 x} = -\sin x$$ Jul 8 '17 at 21:21
• @quasi, if $a=x^2$, isn't $\sqrt{a}=\sqrt{x^2}=\sqrt{(\pm x)^2}=\pm x$?
– MrAP
Jul 8 '17 at 21:28
• @MrAP: No, $\sqrt{x}$ is a function (i.e., single valued). Thus, for example, while it's true that the number $9$ has two square roots, $\pm 3$, the expression $\sqrt{9}$ yields $3$, not $\pm 3$. Jul 8 '17 at 21:31