Compute $\lim_{x\rightarrow 0} \frac{\ln{\cos{x}}}{e^{2x^2}-\cos{x}}$

Question

This should be done without using L'Hopital's rule and without using the $\sec(x)$ and $\csc{x}$ functions. I tried substituting $t=\cos{x}\rightarrow x=\arccos{t}$ and obtain

$$\lim_{t\rightarrow 0} \frac{\ln{(1+t)}}{e^{2\arccos^2(t)}-t},$$

Dividing nominator and denominator by $t$ I get

$$\lim_{t\rightarrow 0} \frac{\frac{\ln{t}}{t}}{\frac{e^{2\arccos^2(t)}}{t}-1}$$

But this gives division by zero. I also tried multilying both sides by $t/t$ but then I get multiplication by zero.

Answer 1

$$\lim_{x\rightarrow 0} \frac{\ln{\cos{x}}}{e^{2x^2}-\cos{x}}=\lim_{x\rightarrow 0} \frac{\ln{(1+\cos{x}-1)}}{(\cos{x}-1)\cdot\frac{e^{2x^2}-1+1-\cos{x}}{2x^2}\cdot\frac{2x^2}{\cos{x}-1}}=\frac{1}{\left(1+\frac{1}{4}\right)\cdot(-4)}=-\frac{1}{5}$$

I used the following. $$\lim_{x\rightarrow0}\frac{\ln(1+x)}{x}=1,$$ $$\lim_{x\rightarrow0}\frac{e^x-1}{x}=1$$ and $$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1.$$