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

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?

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  • $\begingroup$ "Euler's identity" ? Which one ? $\endgroup$ – Yves Daoust Oct 29 '18 at 9:59
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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 $$\frac{1-y}{y^2}= \frac{1}{y^2} - \frac{1}{y} \stackrel{y \to \infty}{\longrightarrow} 0$$

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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 $a>1$

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  • $\begingroup$ 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? $\endgroup$ – Danielle Oct 29 '18 at 6:41
  • $\begingroup$ @Danielle, See oxfordmathcenter.com/drupal7/node/94 $\endgroup$ – lab bhattacharjee Oct 29 '18 at 6:43
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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 $\infty$.

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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 ?

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Obviously,

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

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

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  • $\begingroup$ 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. $\endgroup$ – gimusi Oct 29 '18 at 7:51

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