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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?

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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.

Using the equality

$$1-\cos2y=2\sin^2y$$

we have with $5x=2y$

$$\lim_{x\to0}\frac{1-\cos5x}{x^2}=\lim_{y\to0}\frac{2\sin^2y}{\left(\frac{2y}{5}\right)^2}=\frac{25}{2}\lim_{y\to0}\left(\frac{\sin y}{y}\right)^2=\frac{25}{2}$$

$\langle$ 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.

$$\lim_{x \to 0} \frac{1-\cos 5x}{x^2}\frac{1+\cos 5x}{1+\cos 5x}=\lim_{x \to 0}\frac{1-\cos^25x}{x^2}\frac{1}{1+\cos5x}=$$

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

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

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:

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

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

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