Evaluating $\lim_{x \to 0} \frac{\sin^{10}(2\sin^{10}(3x))}{x^{100}}$ without l'Hospital

Question

I'm quite at a loss. I would 'expect' it to be equal to $1$ since $\lim\limits_{x \to 0}2\sin(3x) = 0$ and $\lim\limits_{x \to 0} \Big (\cfrac{\sin x}{x^{10}} \Big)^{10} = 1?$

Hints please! (Preferably not the full answer)

Answer 1

You want$$\lim_{x\to0}\left(\frac{\sin(2\sin^{10}3x)}{x^{10}}\right)^{10}=\left(\lim_{x\to0}\frac{\sin(2\sin^{10}3x)}{2\sin^{10}3x}\lim_{x\to0}2\cdot 3^{10}\left(\frac{\sin 3x}{3x}\right)^{10}\right)^{10}=\cdots$$

Answer 2

We have $$ \lim_{x \to 0} \frac{\sin^{10}(2\sin^{10}(3x))}{x^{100}} = \left(\lim_{x \to 0} \frac{\sin(2\sin^{10}(3x))}{x^{10}}\right)^{10}, $$ so it suffices to evaluate the limit within the power. Now $$ \lim_{x \to 0} \frac{\sin(2\sin^{10}(3x))}{x^{10}} = \lim_{x \to 0} \frac{\sin(2\sin^{10}(3x))}{2\sin^{10}(3x)} \frac{2\sin^{10}(3x)}{x^{10}}= \lim_{x \to 0} \frac{\sin(2\sin^{10}(3x))}{2\sin^{10}(3x)} \times \lim_{x \to 0}\frac{2\sin^{10}(3x)}{x^{10}}, $$ and the first part of this product is just 1. So we are looking for $$ \lim_{x \to 0}\frac{2\sin^{10}(3x)}{x^{10}} = 2\left(\lim_{x \to 0} \frac{\sin(3x)}{x}\right)^{10}. $$ Finally this innermost limit equals $$ \lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \times 3 = 3. $$ Working back, we see that your answer is $$2^{10}3^{100} = 527746581229579602981336196879996183246958102529024.$$

Answer 3

Look at the leading terms: $$\sin 3x = 3x + \frac{3^3}{3!}x^3 + \cdots$$ $$2\sin^{10}3x = 2(3x + \frac{3^3}{3!}x^3 + \cdots)^{10} = 2\cdot 3^{10}x^{10} + \cdots$$ $$\sin(2\sin^{10}3x) = (2\cdot 3^{10}x^{10} + \cdots) + \frac{1}{3!}(2\cdot 3^{10}x^{10} + \cdots)^3 + \cdots=2\cdot 3^{10}x^{10} + \cdots$$ $$\sin^{10}(2\sin^{10}3x)=(2\cdot 3^{10}x^{10} + \cdots)^{10} = 2^{10}3^{100}x^{100}+\cdots$$ $$\frac{\sin^{10}(2\sin^{10}3x)}{x^{100}}=\frac{2^{10}3^{100}x^{100}+\cdots}{x^{100}}=2^{10}3^{100}+\cdots$$

All of the "$\cdots$" terms involve higher powers of $x$ which will disappear in the limit. So $$\lim\limits_{x\to 0}\frac{\sin^{10}(2\sin^{10}3x)}{x^{100}} = \boxed{2^{10}3^{100}}.$$

Answer 4

As often, the most straightforward way is via equivalents: as $\sin u\sim_0 u$, we have $$\frac{\sin^{10}(2\sin^{10}(3x))}{x^{100}}\sim_0\frac{\sin^{10}(2(3x)^{10})}{x^{100}}\sim_0\frac{\bigl(2(3x)^{10}\bigr)^{10}}{x^{100}}=\frac{2^{10}3^{100}\not x^{100}}{\not x^{100}}=2^{10}3^{100}.$$