Different method to take limit gives different results? Question:


  $$\lim_{x\to0}\frac{\sin(3x^2)}{\log\cos(2x^2-x)}=?$$


My method:

Apply L-H rule, since the form is $\frac 00$:

$$\lim_{x\to0}-\frac{6x\cos(3x^2)\cos(2x^2-x)}{\sin(2x^2-x)\cdot(4x-1)}$$
Putting in the value for the determinate terms we get:
$$\lim_{x\to0}\frac{6x}{\sin(2x^2-x)}$$
Apply L-H rule again to get:
$$\lim_{x\to0}\frac{6}{4x\cos(2x^2-x)}$$

Now, the numerator is $\not\to0$ and the denominator is $\to 0$, thus the limit is undefined! But, the correct answer is not undefined. I don't want to know the correct solution, but I wish to know where I went wrong.
The problem is that after$$\lim_{x\to0}\frac{6x}{\sin(2x^2-x)},$$you should have obtained$$\lim_{x\to0}\frac6{(2x-1)\cos(2x^2-x)}=-6.$$
$$=-6\lim_{x\to0}\dfrac{\sin(3x^2)}{3x^2}\cdot\lim_{x\to0}\dfrac{-\sin^2(2x^2-x)}{\ln(1-\sin^2(2x^2-x))}\cdot\lim_{x\to0}\left(\dfrac{2x^2-x}{\sin(2x^2-x)}\right)^2\cdot\left(\lim_{x\to0}\dfrac x{2x^2-x}\right)^2$$

as $2\ln y=\ln(y^2)$ for $y&gt;0$