Hint on finding limit without application of L'Hospital's rule (mathematical analysis)

I've been trying to solve this limit without application of L'Hospital's rule for some time now, but with no success, tried a couple of approaches but all end up in dead end.

$$\lim_{x\to0} \frac{\ln(e+x)-e^x}{\cos^2x-e^x}$$

any kind of hint would be appreciated.

Answer

$$\dfrac{\ln(x+e)-e^x}{\cos^2x-e^x}$$

$$=\dfrac{1+\ln(1+x/e)-e^x}{(\cos x+e^{x/2})(\cos x-e^{x/2})}$$

$$=\dfrac{-\dfrac1e\cdot\dfrac{\ln(1+x/e)}{x/e}-\dfrac{e^x-1}x}{(\cos x+e^{x/2})\left(-\dfrac12\cdot\dfrac{e^{x/2}-1}{x/2}-\dfrac{1-\cos x}x\right)}$$

Now $\dfrac{1-\cos x}x=\dfrac x{1+\cos x}\cdot\left(\dfrac{\sin x}x\right)^2$