# Question 47ab6

Oct 5, 2017

The limit is 2.

#### Explanation:

L'Hopital's Rule is one way to get the answer. The given limit

${\lim}_{x \to 0} \frac{1 - \cos \left(2 x\right)}{{x}^{2}}$ is a "$\frac{0}{0}$ indeterminate form" (since both the top and bottom approach zero as $x$ approaches zero).

This is a situation where L'Hopital's Rule can be used. Differentiating the top and bottom individually to create a new fraction, the original limit will equal the following limit (if the following limit exists):

${\lim}_{x \to 0} \frac{2 \sin \left(2 x\right)}{2 x} = {\lim}_{x \to 0} \frac{\sin \left(2 x\right)}{x}$.

This is, once again, a $\frac{0}{0}$ indeterminate form. L'Hopital's Rule can again be used to say it will equal the following limit (if the following limit exists):

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

But this limit exists because the function in question is continuous. The limit can be evaluated by substitution: ${\lim}_{x \to 0} 2 \cos \left(2 x\right) = 2 \cos \left(0\right) = 2 \cdot 1 = 2$.

Putting these things together implies that ${\lim}_{x \to 0} \frac{1 - \cos \left(2 x\right)}{{x}^{2}} = 2$.

Another way to get the answer is to use the Taylor (a.k.a. Maclaurin) series for the cosine function expanded about 0. That is, the fact that cos(z)=1-z^2/(2!)+z^4/(4!)-z^6/(6!)+cdots for all $z$.

This implies that (1-cos(2x))/(x^2)=(1-(1-(2x)^2/(2!)+(2x)^4/(4!)-(2x)^6/(6!)+cdots))/(x^2)#

$= \frac{2 {x}^{2} - \frac{2}{3} {x}^{4} + \frac{4}{45} {x}^{6} - \cdots}{{x}^{2}} = 2 - \frac{2}{3} {x}^{2} + \frac{4}{45} {x}^{4} - \cdots$ for all $x \ne 0$.

But the value at $x = 0$ is irrelevant for the value of the original limit (the original function is undefined at 0). We can say that

${\lim}_{x \to 0} \frac{1 - \cos \left(2 x\right)}{{x}^{2}} = {\lim}_{x \to 0} \left(2 - \frac{2}{3} {x}^{2} + \frac{4}{45} {x}^{4} - \cdots\right)$

This last function is continuous at $x = 0$, so the value of this last function at zero is relevant. We can evaluate the limit by substitution of $x = 0$ to get

${\lim}_{x \to 0} \frac{1 - \cos \left(2 x\right)}{{x}^{2}} = {\lim}_{x \to 0} \left(2 - \frac{2}{3} {x}^{2} + \frac{4}{45} {x}^{4} - \cdots\right) = 2 - 0 + 0 - 0 + 0 - \cdots = 2$

Oct 5, 2017

$2.$

#### Explanation:

We use the Standard Limit : ${\lim}_{\theta \to 0} \sin \frac{\theta}{\theta} = 1.$

Recall that, $1 - \cos 2 x = 2 {\sin}^{2} x .$

$\therefore \text{ The Reqd. Lim.=} {\lim}_{x \to 0} \frac{1 - \cos 2 x}{x} ^ 2 ,$

$= \lim \frac{2 {\sin}^{2} x}{x} ^ 2 ,$

$= \lim 2 {\left(\sin \frac{x}{x}\right)}^{2} ,$

$= 2 {\left(1\right)}^{2} ,$

$= 2.$