Limit Evaluation - $\lim_{x\to \infty} \frac{1-e^x}{e^{2x}}$ $\lim_{x\to \infty} \frac{1-e^x}{e^{2x}}$
My guess is to evaluate by dividing all terms by $e^x$, which works and gives me Eulers identity. 
But why should that be right? I thought we are only supposed to divide by the highest exponent term in the denominator? But when I do that, I cannot get a solution? How and when is it ok to divide by an exponent in the numerator?

For $x>0$ we have $e^x>1$ and $e^{x}>x$ (use the power series for $e^x$).
Hence $|\frac{1-e^x}{e^{2x}}| \le \frac{1+e^x}{e^{2x}} \le \frac{2e^x}{e^{2x}}=\frac{2}{e^{x}} \le \frac{1}{x}$.
Can you proceed ?
