Calculating Euler Number limit Please, so far I did
$$\lim_{x\to +\infty}\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}},$$ but I can write
$$\frac{x^2-x+1}{x+2}=1+\frac{x^2-2x-1}{x+2}=1+\frac{1}{\frac{x+2}{x^2-2x-1}}.$$
But
$$\lim_{x\to +\infty}\frac{x+2}{x^2-2x-1}=0,$$ so I can not use 
$$e =\lim_{N\to \infty}(1+\frac{1}{N})^N$$

Let $\displaystyle y=\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}}$ then 
$$\ln y= \frac{1}{x-1}\ln(x^2-x+1)-\frac{1}{x-1}\ln(x+2)$$
$$\lim_{y\rightarrow \infty} \ln y=\lim_{x\rightarrow \infty}\frac{2x-1}{x^2-x+1}-\lim_{x\rightarrow \infty}\frac{1}{x+2}=0,\ L'Hopital$$
Then $\lim_{y\rightarrow \infty}y=1$
