What's wrong with my calculation of a limit here?

Question

So, I have to find the following limit: $\lim _{x\to \:0+}\left(\frac{\left(1-\cos \left(2x\right)\right)^{14}\left(1-\cos \left(7x\right)\right)^2\sin ^{14}\left(9x\right)}{\tan ^{14}\left(x\right)\left(\ln \left(8x+1\right)\right)^{30}}\right)$

I solved it by splitting it into three limits as follows: $$\begin{align} &\lim _{x\to \:0+}\left(\frac{\left(1-\cos \left(2x\right)\right)^{14}\left(1-\cos \left(7x\right)\right)^2\sin ^{14}\left(9x\right)}{\tan ^{14}\left(x\right)\left(\ln \left(8x+1\right)\right)^{30}}\right) \\ &=\lim _{x\to 0+}\frac{\left(1-cos\left(2x\right)\right)^{14}}{\tan ^{14}\left(x\right)}\cdot \frac{\left(1-cos\left(7x\right)\right)^2}{\left(\ln \left(8x+1\right)\right)^2}\cdot \frac{sin\left(9x\right)^{14}}{\left(\ln \left(8x+1\right)\right)^{28}} \\ &=\lim_{x\to 0+}\frac{\left(1-cos\left(2x\right)\right)^{14}}{\tan ^{14}\left(x\right)}\cdot \lim_{x\to\:0+}\frac{\left(1-cos\left(7x\right)\right)^2}{\left(\ln\left(8x+1\right)\right)^2}\cdot\lim_{x\to 0+}\frac{sin\left(9x\right)^{14}}{\left(\ln\left(8x+1\right)\right)^{28}} \\ &=\left(\lim _{x\to \:\:0+}\frac{\left(1-cos\left(2x\right)\right)}{\tan \left(x\right)}\right)^{14}\cdot \left(\lim _{x\to 0+}\frac{\left(1-cos\left(7x\right)\right)}{\:\ln \left(8x+1\right)}\right)^2\cdot \left(\lim _{x\to 0+}\frac{sin\left(9x\right)}{\left(\ln\left(8x+1\right)\right)^2}\right)^{14} \end{align}$$

Using L'Hospital's rule to solve these separate limits, you get $$\left(\lim _{x\to \:\:0+}\frac{\left(1-cos\left(2x\right)\right)}{\tan \left(x\right)}\right)^{14}\cdot \left(\lim _{x\to 0+}\frac{\left(1-cos\left(7x\right)\right)}{\:\ln \left(8x+1\right)}\right)^2\cdot \left(\lim _{x\to 0+}\frac{sin\left(9x\right)}{\left(\ln\left(8x+1\right)\right)^2}\right)^{14} \\ =0^{14}\cdot 0^2\cdot \left(-\frac{9}{16}\right)^{14} \\ =0$$

However, the automated homework system did not accept 0 as the correct answer. What did I do wrong?

Answer 1

Let $a,b>0,c>0$ be non zero constants. Use (dividing by $x$ both denominator and numerator) the follwing:

$$\lim_{x\to 0}\frac{\sin(ax)}{\tan(bx)}=\frac{a}{b}.$$

and you can use l'hôpital to see that:

$$\lim_{x\to 0^{+}}\frac{\ln(cx+1)}{x}=c.$$

To first simplify, (dividing by $x^{14}$ both denominator and numerator)

$$\lim_{x\to 0}\frac{\sin(9x)^{14}}{\tan(x)^{14}}=9^{14}.$$

and also:

$$ \lim_{x\to 0^{+}}\frac{\ln(8x+1)}{x}=8.$$

Your limit will be:

$$\lim _{x\to \:0+}\frac{\left(1-\cos \left(2x\right)\right)^{14}\left(1-\cos \left(7x\right)\right)^2}{(8x)^{30}}$$

Answer 2

Sanity check:

For small $x$, $$1-\cos x=\Theta(x^2),\sin x=\Theta(x),\tan x=\Theta(x),\log(x+1)=\Theta(x).$$

Then the expression is of order

$$\frac{x^{28}x^4x^{14}}{x^{14}x^{30}}=x^2.$$

(Which tends to $0$, of course.)