# Lim x->0 √1-cos 2x/√2x=? Thanks

##### 3 Answers
Jun 27, 2017

${\lim}_{x \to 0} \left(\frac{\sqrt{1 - \cos \left(2 x\right)}}{\sqrt{2 x}}\right) = 0$

#### Explanation:

Disclaimer: I don't know much about limits besides the general idea of what they are, but here's my take on this:

Looking at the limit:
${\lim}_{x \to 0} \left(\frac{\sqrt{1 - \cos \left(2 x\right)}}{\sqrt{2 x}}\right)$
iff sqrt(lim_(x->0) ((1-cos(2x))/(2x))

Using l'hopital's method, we can see that both the numerator and denominator is approaching zero, so:
If
${\lim}_{x \to c} \left(f \left(x\right)\right) = {\lim}_{x \to c} \left(g \left(x\right)\right) = 0 \mathmr{and} \pm \infty$

then ${\lim}_{x \to c} \frac{f \left(x\right)}{g} \left(x\right) = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

In this case $f \left(x\right) = 1 - \cos \left(2 x\right)$, $f ' \left(x\right) = 2 \sin \left(2 x\right)$
$g \left(x\right) = 2 x$, $g ' \left(x\right) = 2$

Therefore:
sqrt(lim_(x->0) ((1-cos(2x))/(2x)) = sqrt(lim_(x->0) ((2sin(2x))/2)

Now we have eliminated the problem of dividing by zero, so
sqrt(lim_(x->0) ((2sin(2x))/2) $= \sqrt{\frac{0}{2}} = 0$

Jun 28, 2017

${\lim}_{x \rightarrow 0} \frac{\sqrt{1 - \cos 2 x}}{\sqrt{2 x}} = \sqrt{{\lim}_{x \rightarrow 0} \frac{1 - \cos 2 x}{2 x}}$

Knowing that the Maclaurin series for $\cos x$ is cosx=sum_(n=0)^oo((-1)^nx^(2n))/((2n)!)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+..., we can see that cos2x=sum_(n=0)^oo((-1)^n(2x)^(2n))/((2n)!)=1-(4x^2)/(2!)+(16x^4)/(4!)-(64x^6)/(6!)+...

Then, we can say the limit is:

=sqrt(lim_(xrarr0)(1-(1-(4x^2)/(2!)+(16x^4)/(4!)-(64x^6)/(6!)+...))/(2x))

=sqrt(lim_(xrarr0)((4x^2)/(2!)-(16x^4)/(4!)+(64x^6)/(6!)+...)/(2x))

=sqrt(lim_(xrarr0)(2x)/(2!)-(8x^3)/(4!)+(32x^6)/(6!)+...)

As the series continues infinitely, all the terms contain an $x$ and will become $0$ as $x \rightarrow 0$.

$= \sqrt{0}$

$= 0$

Jun 28, 2017

${\lim}_{x \to 0} \frac{\sqrt{1 - \cos \left(2 x\right)}}{\sqrt{2 x}} = 0$

#### Explanation:

${\lim}_{x \to 0} \frac{\sqrt{1 - \cos \left(2 x\right)}}{\sqrt{2 x}} = {\lim}_{x \to 0} \sqrt{\frac{1 - \cos \left(2 x\right)}{2 x}}$

Using the substitution $\theta = 2 x$, when $x$ goes to $0$, $\theta$ also goes to $0$

${\lim}_{x \to 0} \sqrt{\frac{1 - \cos \left(\theta\right)}{\theta}}$

Which, if you look closely, you'll see is the square root of a fundamental limit, one we know to be 0. So the limit as a whole is $0$

(If you don't quite recognize it as a fundamental limit, you can tweak it so it involves the other fundamental limit ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$. A non-circular proof of that, however, requires either to use the epsilon-delta definition of limits or some geometric arguments)