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