Evaluate without L'Hospital's rule: $\displaystyle\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}$
My attempt:
$$e^{x^2}-\cos x=e^{x^2}-1+1-\cos x=x^2\cdot\left(\frac{e^{x^2}-1}{x^2}+\frac{1-\cos x}{x^2}\right)$$ $$\frac{\ln(e+x^2)-\cos x}{x^2}=\frac{\ln(e+x^2)-1+1-\cos x}{x^2}=\frac{\ln\left(1+\frac{x^2}{e}\right)}{\frac{x^2}{e}\cdot e}+\frac{1-\cos x}{x^2}$$ $$\lim_{x\to 0}\frac{\ln(e+x^2)-\cos x}{e^{x^2}-\cos x}=\lim_{x\to 0}\frac{\frac{\ln\left(1+\frac{x^2}{e}\right)}{\frac{x^2}{e}}\cdot\frac{1}{e}+\frac{1-\cos x}{x^2}}{\frac{e^{x^2}-1}{x^2}+\frac{1-\cos x}{x^2}}=\frac{\frac{1}{e}+\frac{1}{2}}{\frac{3}{2}}=\frac{2(2+e)}{6e}$$
$$L=\lim_{x \rightarrow 0} \frac{\ln (e+x^2)-\cos x}{e^{x^2}-\cos x}$$ Let us use $$e^{x^2}=1+x^2+O(x^4), \cos x= 1+ x^2/2+O(x^4), \ln(1+z)=z-z^2/2+O(z^4),$$ then $$L=\lim_{x \rightarrow 0}\frac{1+x^2/e-1+x^2/2+O(x^4)}{3x^2/2+O(x^4)}=\frac{2+e}{3e}$$