How to find the limit of $\frac{1−\cos 5x}{x^2}$ as $x\to 0$?

What is the right approach to calculate the Limit of $(1-\cos(5x))/x^2$ as $x \rightarrow 0$?

From Wolfram Alpha, I found that: $$\lim_{x \to 0} \frac{1- \cos 5x}{x^2} = \frac{25}{2}.$$ How do I get that answer?

• Please verify formatting didn't change your question. Nov 14, 2013 at 20:49

Using the equality $$1-\cos2y=2\sin^2y$$ we have with $5x=2y$ $$\lim_{x\to0}\frac{1-\cos5x}{x^2}=\lim_{y\to0}\frac{2\sin^2y}{\left(\frac{2y}{5}\right)^2}=\frac{25}{2}\lim_{y\to0}\left(\frac{\sin y}{y}\right)^2=\frac{25}{2}$$

$$\lim_{x \to 0} \frac{1-\cos 5x}{x^2}\frac{1+\cos 5x}{1+\cos 5x}=\lim_{x \to 0}\frac{1-\cos^25x}{x^2}\frac{1}{1+\cos5x}=$$ $$=\lim_{x\to 0}\left(\frac{\sin5x}{5x}\right)^2\frac{25}{1+\cos5x}=\frac{25}{2}$$

• +1 I just saw this question after this question was marked as a duplicate. This answer is almost identical to mine there.
– robjohn
Jul 18, 2016 at 17:50

$\langle$ Begin of rant $\rangle$ It seems to be an effect of the choice made in some educational systems to inflate the importance of l'Hôpital's rule, that this (at best anecdotal) trick becomes the alpha and the omega of the approach by several MSEers of every question involving the local analysis of a real function. This happens at the expense of several other tools, often more important mathematically and at least as easy to use. As a consequence, let me humbly suggest that the extent of this monopole might be completely disproportionate, and even detrimental to the learning process of the field. $\langle$ End of rant $\rangle$

In the present case, a direct approach uses a quite rudimentary limited expansion of the cosine function at zero, namely, the fact that $$1-\cos(u)\sim\tfrac12u^2,\qquad u\to0.$$ For $u=5x$, the fact that the limit asked about is $\frac{25}2$ becomes obvious.

To come back to the object of the rant, is it unreasonable to ask that anybody interested in mathematical analysis might know the equivalent used above (saying that the graph of cosine looks like a parabola around $0$), and that $\sin(u)\sim u$ and $\tan(u)\sim u$, and a few other similar expansions, say, $\mathrm e^u-1\sim u$ (fundamental) and $\sqrt{1+u}-1\sim\frac12u$ (each saying what is the slope of the tangent to a graph at $0$), rather than the intricacies of a black box called l'Hôpital's rule (whose requirements, by the way, are almost never checked)? Honestly, I do not think so.

• I wholeheartedly agree with the first and the last paragraph. I also concur that the solution of the problem in between is the best and easiest solution of this problem. Nov 14, 2013 at 21:17
• @hhsaffar Really? Please test the "power" of l'H on the limit of $(1-\cos(5x)+\sin(17x)^3-\mathrm e^{-42/x^2})/(x^2+\sqrt[3]{x^7+x^9})$ when $x\to0$.
– Did
Nov 14, 2013 at 21:19
• @Did I have nothing against knowing the limited expansions, however, putting the instance you proposed aside, with l'H you have to memorize much less, you can know nothing about expansions and series, a topic that is not frequently taught in high schools, and yet you can answer almost all high school level\early college questions.It is powerful because it lets you to solve lots of problems without knowing much. Nov 14, 2013 at 21:26
• @Did : not everyone will be able or willing to memorize the power series expansions of all the functions you think are important. What about $\tan^{-1}$? $\ln$? Where do you draw the line? L'Hopital's Rule is easy to remember and use (in typical problems, there are no "intricacies"). In addition, few students know what "little-o" notation means. It, as well as "big-O" notation, are not part of most calculus sequences. Whether they should be is a separate question. I do think many students overuse it, and attempt to use it for everything, before trying a little algebra and common sense. Nov 15, 2013 at 2:32
• Actually, an interesting question would be to determine when the place of l'Hôpital's rule began to be so preeminent in certain curricula, which other, previous, expertises it replaced, and how the move was justified (if it was perceived) at the time by its proponents.
– Did
Nov 15, 2013 at 2:48

We can use L’Hôpital’s rule, where $f(x) = 1-\cos 5x$ and $g(x)=x^2$, as $$\lim_{x \to 0} (1-\cos 5x) = \lim_{x \to 0} x^2 = 0,$$ and $g'(x) = 2x \neq 0$ for all $x\in\mathbb{R}\backslash\{0\}$.

Then we get $$\lim_{x \to 0} \frac{1-\cos 5x}{x^2} = \lim_{x \to 0} \frac{5\sin 5x}{2x} = \lim_{x \to 0} \frac{25 \cos 5x}{2} = \frac{25}{2},$$ where we apply L’Hôpital’s rule twice.

• It should be $\lim_{x \to 0} \dfrac{1-\cos 5x}{x^2} = \lim_{x \to 0} \dfrac{5\sin 5x}{\mathbf{2}x}$ Nov 14, 2013 at 21:06
• @K.Rmth: of course, fixed. Thanks. Nov 14, 2013 at 21:08
• $$\lim_{x \to 0} \frac{5\sin 5x}{2x} = \frac{5}{2} \lim_{x \to 0} \frac{\sin 5x}{x}$$ is just the definition of the derivative of $\sin(5x)$ at $0$... So the second L'H use is circular logic.... Nov 14, 2013 at 21:27

$\cos(2\alpha) = 1-2\sin ^2 \alpha$

Therefore

$\frac{1-\cos (5x)}{x^2}=\frac{2 \sin ^2 \left( \frac{5x}{2}\right)}{x^2} = \frac{25}{2} \left( \frac{\sin \frac{5}{2}x}{\frac{5}{2}x} \right)^2 \rightarrow \frac{25}{2}$

Use L'Hopital's Rule:

Since the limit evaluates to the form $\frac 00$ as is, we can take the derivative of the numerator and denominator, and then evaluate the limit as $x\to 0$; if it still evaluates to the form $\frac 00$, take the derivative of the numerator and denominator again, and evaluate the resulting limit as $x\to 0$. Etc: Here, we apply L'H twice:

$$\lim_{x \to 0} \frac{1-\cos 5x}{x^2} \quad \overset{L'H}{=} \quad \lim_{x \to 0} \frac{5\sin 5x}{2x}\quad \overset{L'H}{=}\quad \lim_{x \to 0} \frac{25 \cos 5x}{2} \quad = \quad\frac{25}{2}$$

• It should be $\lim_{x \to 0} \dfrac{1-\cos 5x}{x^2} = \lim_{x \to 0} \dfrac{5\sin 5x}{\mathbf{2}x}$ Nov 14, 2013 at 21:07

There is a really easy way of doing this, take derivatives. Namely: $$\partial_{5} \frac{1- Cos(5x)}{x^{2}} =\frac{xSin(5x)}{x^{2}}=\frac{Sin(5x)}{x} \to 5$$ Now, since we took a derivative w.r.t 5, we must take the integral with respect to 5: $$\int 5 \mathrm{d}5=\frac{5^2}{2}=25/2$$

L'Hospital's Rule, twice. You will have a $25cos5x$ in the num, and a 2 in the denom, supporting your answer