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What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$? My calculator tells me the answer is -9/2, but I don't know how to solve without substituting values of x. I suspect there is some trigonometric identity involved.

CAGT
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    It depends on what is in your bag of tricks. Since $\cos(3x) = 1-{1 \over 2!} (3x)^2+{1 \over 4!} (3x)^4- \cdots$, we see that the above is equal to ${-{1 \over 2!} (3x)^2+{1 \over 4!} (3x)^4- \cdots \over x^2}$ and taking limits as $x \to 0$ gives $- { 3^2 \over 2!}$. – copper.hat Feb 16 '16 at 05:59
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    Multiply top and bottom by $\cos(3x)+1$. We get $-\frac{1}{\cos (3x)+1}\cdot\frac{\sin^2 (3x)}{x^2}$. If things are not yet clear, rewrite as $-\frac{9}{\cos (3x)+1}\cdot\frac{\sin^2 (3x)}{(3x)^2}$. – André Nicolas Feb 16 '16 at 06:00

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From the trigonometric identity $\;\color{blue}{\cos(2\theta)=1-2\sin^2\theta}\;$ we have \begin{align} \lim_{x\to 0}\frac{\cos(3x)-1}{x^2}&=\lim_{x\to 0}\frac{1-2\sin^2(3x/2)-1}{x^2}\\ &=\lim_{x\to 0}\frac{-2\sin^2(3x/2)}{x^2}\\ &=-2\cdot\frac{9}{4}\cdot\lim_{x\to 0}\frac{\sin^2(3x/2)}{\frac{9}{4}x^2}\\ &=-2\cdot\frac{9}{4}\cdot\left(\lim_{x\to 0}\frac{\sin(3x/2)}{3x/2}\right)^2\\ &=-2\cdot\frac{9}{4}\cdot\left(1\right)^2\\ &=\boxed{\color{blue}{-\frac{9}{2}}} \end{align}

Ángel Mario Gallegos
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$$\lim_{x\rightarrow 0}{\cos(x)}\sim 1-\frac{x^2}{2}$$

That's a formula . Derived from the below expression.

$$\cos(x) = 1-{1 \over 2!} (x)^2+{1 \over 4!} (x)^4- \ldots$$

Now, use it and get the answer.

$$\cos(3x) = 1-\frac{9x^2}{2}$$

Win Vineeth
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    This is not quite correct. You can't have a formula where the bound variable is $x$ on one side but is unbound on the other. – copper.hat Feb 16 '16 at 06:00
  • @Win Vineeth where can I find such formulas? – CAGT Feb 16 '16 at 06:00
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    It is impossible $\lim_{x\to 0}\cos(x)=1-\frac{x^2}{2}$. I suppose you had a type error – sinbadh Feb 16 '16 at 06:01
  • @CAGT there is a comment on your question. It is derived from that. – Win Vineeth Feb 16 '16 at 06:01
  • @sinbadh $\cos(3x) = 1-{1 \over 2!} (3x)^2+{1 \over 4!} (3x)^4- \cdots$ It is derived from that. – Win Vineeth Feb 16 '16 at 06:01
  • No @WinVineeth. It is not true. $\lim_{x\to 0}\cos(x)=1$ (a number), while $1-\frac{x^2}{2}$ is a function – sinbadh Feb 16 '16 at 06:03
  • Thanks guys, definitely need a refresh on many identities here. – CAGT Feb 16 '16 at 06:03
  • @sinbadh Yes it is a function. $\lim_{x\to 0}\sin(x)/x=1$ is sometimes shown as $\lim_{x\to 0}\sin(x)=x$ It is an approximation. – Win Vineeth Feb 16 '16 at 06:04
  • No. it uses something like $\sim$, not $=$. Is for that i supposed you had a typing error – sinbadh Feb 16 '16 at 06:07
  • Indeed, note that exactly the same observation was made by @cooper.hat – sinbadh Feb 16 '16 at 06:08
  • @sinbadh Yes.. you are right. I edited. Thank you. I was just not that careful.. – Win Vineeth Feb 16 '16 at 06:08
  • @WinVineeth: It would be better to write $\cos x \approx 1-{ 1\over 2!} x^2 + \cdots$. – copper.hat Feb 16 '16 at 06:09
  • @copper.hat Is it? I was taught that it was exactly equal to that particular infinite series.. – Win Vineeth Feb 16 '16 at 06:10
  • No @WinVineeth. In that case, it is an equality cause you are talking about the complete infinite series. On the other hand, a way to express the particular idea would be $\cos x\sim 1-\frac{x^2}{2!}$ near of $0$ – sinbadh Feb 16 '16 at 06:10
  • @WinVineeth: I'm not sure if you are being facetious or not. Assuming not, what I mean is that when you have $\lim_{x \to 0} (\text{something})$ then all the $x$s in $\text{something}$ 'disappear', so you have a constant on the left hand side and a non constant function of $x$ on the right hand side which is incorrect. It is true that $\cos x$ is given by the infinite Taylor series, the $\approx$ is just ignoring terms of order $x^3$ and above (in this example). – copper.hat Feb 16 '16 at 06:13
  • @copper.hat Oh.. Now I understand what you meant. Ya, that's right..and no, I wasn't being facetious. Thank you. – Win Vineeth Feb 16 '16 at 06:16
  • Taylor's? really? :| – BCLC Feb 16 '16 at 10:37
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Top and bottom both go to zero, so use L'Hospitals' rule, take the derivative of the top and bottom, and you get $\frac{-3\sin(3x)}{2x}$. Top and bottom still go to zero so take derivatives again to get $\frac{-9\cos(3x)}{2}$ Now the limit is $-9/2$.

Gregory Grant
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As Andre Nicolas already said in comment, you can do algebraic manipulations to get the limit

If you are not familiar with Taylor expansion of $\cos x$, which are easily applied here, you have to somehow manage to remove the terms which cause the trouble:

We divide and multiply by $\cos(3x) +1$ and try to convert it into a function of $\sin$

$$\frac{\cos(3x)-1}{x^2}\times\frac{\cos(3x)+1}{\cos(3x)+1}$$

$$= \frac{-\sin^2 3x}{x^2 (\cos(3x) + 1)}$$

Now we must do some adjustments to use: $\lim_{x\to 0}\dfrac{\sin x}{x} = 1$

So as $x\to 0$, $3x \to 0$. WE need $3^2$ in the numerator, so divide and multiply by $3^2$

$$=\frac{-9}{\cos(3x)+1}\times \left(\frac{\sin 3x}{3x}\right)^2=-\frac{9}{2}$$

Max Payne
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$$\lim_{x \to 0}\frac{\cos(3x)-1}{x^2}=$$ $$-9 \cdot \lim_{x\to 0} \frac{1-\cos(3x)}{(3x)^2} \stackrel{t=3x}{=} \ldots$$

Lonidard
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