Limit of $\ln \cos(2x^2-x)$

Question

Find $$\lim_{x \to 0}\frac{\sin(6x^2)}{\ln \cos(2x^2-x)}$$ I can write the numerator using series expansion, what about denominator?

Answer 1

Hint

Start with the series expansion of $\cos(y)$ around $y=0$; in the result, replace $y$ by $2x^2-x$. You should then arrive to $$\cos(2x^2-x)=1-\frac{x^2}{2}+2 x^3+O\left(x^4\right)$$ Now, consider the series expansion of $\log(1+y)$. In the result, replace $y$ by $-\frac{x^2}{2}+2 x^3$. This will give you the expansion of the denominator.

I am sure that you can take from here.

Answer 2

We use the fundamental trigonometric and exponential limits:

$$\lim_{x\to0}\color{red}{\frac{\sin(x)}{x}}=1,\qquad\lim_{x\to0}\color{blue}{\frac{\cos(x)-1}{x^2/2}}=-1,\qquad\lim_{x\to0}\color{green}{\frac{\ln(1+x)}{x}}=1.$$

$$\begin{align}\lim_{x \to 0}\frac{\sin(6x^2)}{\ln \cos(2x^2-x)}&=\lim_{x \to 0}\color{red}{\frac{\sin(6x^2)}{6x^2}}\frac{6x^2}{\ln \cos(2x^2-x)}\\&=\lim_{x\to0}\frac{6x^2}{\ln \cos(2x^2-x)}\\&=\lim_{x\to0}\color{green}{\frac{\cos(2x^2-x)-1}{\ln\left(1+ \cos(2x^2-x)-1\right)}}\color{blue}{\frac{(2x^2-x)^2/2}{\cos(2x^2-x)-1}}\frac{6x^2}{(2x^2-x)^2/2}\\&=-\lim_{x\to0}\frac{6x^2}{(2x^2-x)^2/2}\\&=-\lim_{x\to0}\frac{6}{(2x-1)^2/2}=-12\end{align}$$

Answer 3

$$\lim_{x\to0}\frac{\sin(6x^2)}{6x^2}\cdot\lim_{x\to0}\frac{\cos(2x^2-x)-1}{\ln[1+\cos(2x^2-x)-1]}\cdot\lim_{x\to0}\frac{6x^2}{\cos(2x^2-x)-1}$$

First two limits converges to $1$

$$\lim_{x\to0}\frac{6x^2}{\cos(2x^2-x)-1}=-\lim_{x\to0}\frac{6x^2[\cos(2x^2-x)+1]}{\sin^2(2x^2-x)}$$

$$=-\lim_{x\to0}[\cos(2x^2-x)+1]\cdot\frac1{\left(\dfrac{\sin(2x^2-x)}{2x^2-x}\right)^2}\cdot\lim_{x\to0}\frac{6x^2}{(2x^2-x)^2}$$

$$\lim_{x\to0}\frac{6x^2}{(2x^2-x)^2}=\lim_{x\to0}\frac{6x^2}{x^2(2x-1)^2}=\frac6{(-1)^2}=\cdots$$

Answer 4

Expand $$\cos(2x^2-x) = 1-\frac12(2x^2-x)^2+O(x^4)$$

Then

$$\ln \cos(2x^2-x)=-\frac12(2x^2-x)^2+O(x^4)$$

Finally,

$$\begin{align} \frac{\sin(6x^2)}{\ln \cos(2x^2-x)}&=\frac{6x^2+O(x^6)}{-\frac12(2x^2-x)^2+O(x^4)}\\\\ &=\frac{-12x^2+O(x^6)}{x^2+O(x^4)} \end{align}$$

which tends to $-12$ as $x$ approaches $0$.