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?

Your tought is correct but there is not needing of Euler'e identity, indeed dividing both numerator and denominator by $e^x$, we obtain
$$\dfrac{1-e^x}{e^{2x}}=\dfrac{\frac1{e^x}-\frac{e^x}{e^x}}{\frac{e^{2x}}{e^x}}=\dfrac{\frac1{e^x}-1}{e^x}$$
and then it suffices to observe that the numerator tends to $-1$ (that is bounded) and the denominator diverges to $\infty$.
