# 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?

• "Euler's identity" ? Which one ? – Yves Daoust Oct 29 '18 at 9:59

## 6 Answers

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

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 $$a>1$$

• Can you explain how you split the limit up here? I don't understand where you got ...$-\lim_{x\to \infty} \frac{1}{e^x}$ from? – Danielle Oct 29 '18 at 6:41
• @Danielle, See oxfordmathcenter.com/drupal7/node/94 – lab bhattacharjee Oct 29 '18 at 6:43

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 $$\infty$$.

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 ?

Obviously,

$$\frac{1-e^x}{e^{2x}}=\frac1{e^{2x}}-\frac1{e^x}$$ tends to $$0$$.

By using Stolz cesaro theorem $$\lim_{x\to \infty}\frac{1-e^x}{e^{2x}}=\lim_{x\to \infty}\frac{1-e^{x+1}-1+e^x}{e^{2(x+1)}-e^{2x}}=\lim_{x\to \infty}\frac{e^x(1-e)}{e^{2x}(e^2-1)}=\lim_{x\to \infty}\frac{1-e}{e^x(e^2-1)}=0$$

• Stolz-Cesaro is aimed for limit of sequences therfore to use that we should evaluate $\lim_{n\to \infty}\frac{1-e^n}{e^{2n}}=L$ and then show that this implies $\lim_{x\to \infty}\frac{1-e^x}{e^{2x}}=L$. For limit of functions a more natural choice is use l'Hopital rule but in that case is really unnecessary. – gimusi Oct 29 '18 at 7:51