Limit Evaluation – $\lim_{x\to \infty} \frac{1-e^x}{e^{2x}}$

Question

$\lim_{x\to \infty} \frac{1-e^x}{e^{2x}}$

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?

Answer 1

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 $$\frac{1-y}{y^2}= \frac{1}{y^2} - \frac{1}{y} \stackrel{y \to \infty}{\longrightarrow} 0$$