Limit $\lim_{x\to 0^{+}}\frac{2(1-\cos(7ax))}{7ax^2}$

Question

$$\lim_{x\to 0^{+}}\frac{2(1-\cos(7ax))}{7ax^2}$$

Is there a solution using trig identities?

Answer 1

$\begin{split}\lim\limits_{x\to0^+}\frac{2(1-\cos7ax)}{7ax^2} &= \lim\limits_{x\to0^+}\frac{4\sin^2(\frac{7ax}{2})}{7ax^2}\\ &=\lim\limits_{x\to0^+}7a\frac{\sin^2(\frac{7ax}{2})}{(\frac{7ax}{2})^2}\\ &=7a \end{split}$

Answer 2

Hint: Write $2t = 7ax$ (i.e. $t =7ax/2$). Then you have: $$\lim_{x\to 0^{+}}\frac{2(1-\cos(7ax))}{7ax^2}={2\over 7a}\lim_{t\to 0^{+}}\frac{49a^2(1-\cos(2t))}{4t^2}$$

$$ ={7a\over 2}\lim_{t\to 0^{+}}\frac{2\sin^2t}{t^2}={7a}$$

Answer 3

With high-school tools: $$\frac{2(1-\cos 7ax )}{7ax^2}=\frac{2(1-\cos^2 7ax )}{7ax^2(1+\cos 7ax)}=\frac{2\cdot 7a}{\underbrace{1+\cos 7ax}_{\substack{\downarrow\\7a}}}\biggl(\underbrace{\frac{\sin 7ax}{7ax}}_{\substack{\downarrow\\1}}\biggr)^2$$

Addendum:

It is also very simple with the more sophisticated notion of equivalent functions: we have this basic equivalence: $\enspace1-\cos u \sim_0\dfrac{u^2}2,$ whence, by the usual rules of asymptotic calculus, $$\frac{2(1-\cos 7ax )}{7ax^2}\sim_0\frac{2}{7ax^2}\frac{(7ax)^2}{2}=7a.$$