I was given the following exercise in a quiz: $$\lim\limits_{x \to 0} \frac{\sin^2(5x)-\sin^2(x)}{x \sin(7x)}$$ How do I even begin? I thought about de Moivre's (decomposition into sum of $sin (x)$ to different powers), but isn't it an overkill here? I looked for ways to reach the limit $\lim\limits_{x \to 0} \frac{\sin x}{x}$ but couldn't find any.
Asked
Active
Viewed 92 times
-2
-
1Taylor series is the way to go if you cannot find a trick to use. A nice trick is also to divide top and bottom by $x^2$ as in here. – Toby Mak Dec 17 '21 at 13:14
-
2Surely you mean L'Hopital's rule, not de Moivre's? – Toby Mak Dec 17 '21 at 13:17
-
https://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b – lab bhattacharjee Dec 17 '21 at 13:46
2 Answers
6
$$\lim_{x \to 0} \dfrac{\sin^2(5x)-\sin^2(x)}{x \sin(7x)} = \lim_{x \to 0} \dfrac{25\left[\dfrac{\sin(5x)}{(5x)}\right]^2-\left[\dfrac{\sin(x)}{x}\right]^2}{7\dfrac{\sin(7x)}{7x}}$$
Could you complete now?
19aksh
- 12,768
-
1Wow. So simple... :(. By limit arithmetics you apply the limit to the numerator and denominator and using the limit presented below by mathcounterexamples.net, you get 24/7. Thank you. – Friedman Dec 17 '21 at 13:33
-
0
Hint
$$\sin^2(5x)-\sin^2(x)= (\sin(5x)-\sin x)(\sin(5x)+\sin x))$$
$$\sin a - \sin b = \dots, \sin a + \sin b = \dots$$
$$\lim_{y \to 0} \frac{\sin y}{y} = 1$$
mathcounterexamples.net
- 70,018