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?
 A: You might see the limit more clearly by substituting


*

*$y= e^x \Rightarrow \lim_{x\to \infty} \frac{1-e^x}{e^{2x}} = \lim_{y\to \infty} \frac{1-y}{y^2}$
So, you get
$$\frac{1-y}{y^2}= \frac{1}{y^2} - \frac{1}{y} \stackrel{y \to \infty}{\longrightarrow} 0$$
A: Why do we need Euler Identity?
$$\lim_{x\to\infty}\dfrac{1-e^x}{e^{2x}}=\left(\lim_{x\to\infty}\dfrac1{e^x}\right)^2-\lim_{x\to\infty}\dfrac1{e^x}=?$$
Now $\lim_{x\to\infty}a^x=\infty$ for $a>1$
A: 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 $\infty$.
A: 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 ?
A: Obviously,
$$\frac{1-e^x}{e^{2x}}=\frac1{e^{2x}}-\frac1{e^x}$$ tends to $0$.
A: By using Stolz cesaro theorem $\lim_{x\to \infty}\frac{1-e^x}{e^{2x}}=\lim_{x\to \infty}\frac{1-e^{x+1}-1+e^x}{e^{2(x+1)}-e^{2x}}=\lim_{x\to \infty}\frac{e^x(1-e)}{e^{2x}(e^2-1)}=\lim_{x\to \infty}\frac{1-e}{e^x(e^2-1)}=0$
