Different method to take limit gives different results?

Question

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.

Answer 1

The problem is that after$$\lim_{x\to0}\frac{6x}{\sin(2x^2-x)},$$you should have obtained$$\lim_{x\to0}\frac6{(2x-1)\cos(2x^2-x)}=-6.$$

Answer 2

$$=-6\lim_{x\to0}\dfrac{\sin(3x^2)}{3x^2}\cdot\lim_{x\to0}\dfrac{-\sin^2(2x^2-x)}{\ln(1-\sin^2(2x^2-x))}\cdot\lim_{x\to0}\left(\dfrac{2x^2-x}{\sin(2x^2-x)}\right)^2\cdot\left(\lim_{x\to0}\dfrac x{2x^2-x}\right)^2$$

as $2\ln y=\ln(y^2)$ for $y>0$

Answer 3

You cannot substitute $x$ by $0$ for just SOME $x$. Either you get a definite value when substituting all $x$ or you don't, in which case you have to try another method (like using the L-H-rule again, on the difficult expression without partial substitution).

Answer 4

Alternatively: $$\lim_\limits{x\to 0} \frac{6x}{\sin (2x^2-x)}=\lim_\limits{x\to 0} \frac{2x^2-x}{\sin (2x^2-x)}\cdot \frac{6x}{2x^2-x}=1\cdot \frac{6}{0-1}=-6.$$