3

So I was again looking through the homepage of YouTube to see if there were any math equations that I thought that I would be able to solve, and I got lucky and found this equation by Cipher$$\text{Evaluate }\lim_{x\to0}\frac{2-\cos(2x)-\cos(4x)}{x^2}$$which I thought I might be able to solve. Here is my attempt at evaluating the limit:$$\text{Since I know that plugging in }0\text{ for }x\text{ will get me}$$$$\lim_{x=0}\frac{2-\cos(0)-\cos(0)}{0^2}\text{ }\longrightarrow\text{ }\frac{2-1-1}{0}\text{ }\longrightarrow\text{ }\frac{0}{0}$$$$\text{which is undefined, so I need to take the left hand and right hand limits}$$$$\text{Right hand limit: }\lim_{x\to0^+}\frac{2-\cos(2x)-\cos(4x)}{x^2}$$

$x$ $y$
$0.01$ $\approx8$
$0.001$ $\approx9$
$0.0001$ $\approx10$
$0.00001$ $\approx10$

Now, taking the left hand limit$$\lim_{x\to0^-}\frac{2-\cos(2x)-\cos(4x)}{x^2}$$

$x$ $y$
$-0.01$ $\approx8$
$-0.001$ $\approx9$
$-0.0001$ $\approx10$
$-0.00001$ $\approx10$

We see that we get the same thing, therefore, the limit of $$\frac{2-\cos(2x)-\cos(4x)}{x^2}$$as $x$ approaches $0$, although undefined at $0$, is $\approx10$.$$\text{My question}$$

Is my solution correct, or is there anything that I could do to evaluate the limit more easily? I am only looking to see if the overall evaluation is correct.

CrSb0001
  • 2,532
  • 1
    Do you know the Taylor expansion of $\cos$ around zero? – Marc-André Brochu Apr 27 '23 at 15:25
  • @Marc-AndréBrochu I'm not really that familiar with it, no – CrSb0001 Apr 27 '23 at 15:26
  • 1
    It gives you an approximation of $\cos$ close to zero. Taylor series are really useful for this kind of problem. In this case, $\cos(x)=1-x^2/2+r$, where $r$ is a quantity that is "negligible" w.r.t. $x^2$, in the sense that you can ignore it in your question because you only divide by $x^2$ in your limit. Try to replace $\cos(x)$ by $1-x^2/2$ in your equation and you'll see what I mean. – Marc-André Brochu Apr 27 '23 at 15:29
  • @Marc-AndréBrochu Thanks, but how would I replace $\cos(2x)$ and $\cos(4x)$? – CrSb0001 Apr 27 '23 at 15:33
  • 1
    Just replace $x$ by $2x$ and $4x$ respectively in the equation I wrote down. – Marc-André Brochu Apr 27 '23 at 15:38

4 Answers4

3

Your numerical evaluation of the limit is pointing towards the correct solution of $10$. To obtain the limit note that as $x\to 0$ $$ \cos x\sim1-\frac{x^2}{2}+\mathcal O(x^4). $$ So as $x\to 0$: $$ \frac{2-\cos 2x-\cos 4x}{x^2}\sim\frac{2-1+\frac{(2x)^2}{2}-1+\frac{(4x)^2}{2}+\mathcal O(x^4)}{x^2}=10+\mathcal O(x^2). $$ Taking the limit, the $\mathcal O(x^2)$ term vanishes giving $$ \lim_{x\to0}\frac{2-\cos 2x-\cos 4x}{x^2}=10. $$

Aaron Hendrickson
  • 6,042
3

Without resorting to series expansions:

$$\begin{align*} & \lim_{x\to0}\frac{2-\cos(2x)-\cos(4x)}{x^2} \\ &= \lim_{x\to0}\frac{2-(1-2\sin^2(x))-(1-2\sin^2(2x))}{x^2} \\ &= 2 \lim_{x\to0}\frac{\sin^2(x)+\sin^2(2x)}{x^2} \\ &= 2 \left(\left(\lim_{x\to0}\frac{\sin(x)}x\right)^2 + 4 \left(\lim_{x\to0} \frac{\sin(2x)}{2x}\right)^2\right) \end{align*}$$

The remaining limit should be familiar.

user170231
  • 19,334
  • Indeed series expansions in this case is really overkilling and useless! – user Apr 27 '23 at 16:02
  • 1
    Nice! Just an observation that we can also use $\frac{\sin^2(x)+\sin^2(2x)}{x^2}= \frac{\sin^2(x)(1+4\cos^2(x))}{x^2}$ – Vasili Apr 27 '23 at 16:10
2

Your guess is correct, for a complete derivation we can use the standard limit

$$\frac{1-\cos x}{x^2} \to \frac12$$

to obtain

$$\frac{2-\cos(2x)-\cos(4x)}{x^2}=4\cdot\frac{1-\cos(2x)}{4x^2}+16\cdot\frac{1-\cos(4x)}{16x^2} \to 2+8=10$$

The given standard limit can be obtained from the well known result

$$\frac{\sin x}{x} \to 1$$

indeed

$$\frac{1-\cos x}{x^2}=\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}=\frac{\sin^2 x}{x^2}\cdot\frac{1}{1+\cos x} \to 1 \cdot \frac12 =\frac12$$

user
  • 154,566
0

Another way by $\cos (2\theta)=1-2\sin^2(2\theta)$ and standard limits

$$\frac{2-\cos(2x)-\cos(4x)}{x^2}=\frac{1-\cos(2x)+2\sin^2(2x)}{x^2}=$$

$$=\frac{1-\cos(2x)}{x^2}+2\cdot \frac{\sin^2(2x)}{x^2}=4\cdot \frac{1-\cos(2x)}{(2x)^2}+8\cdot\frac{\sin^2(2x)}{(2x)^2} \to 4\cdot \frac12+8\cdot 1=10$$

user
  • 154,566