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Given $$\lim_{x\to 0} (e^x-2x)^\frac{1}{x} $$ I know that you take the natural log $$\lim_{x\to 0} \frac{1}{x}\ln(e^x-2x) $$ which is $$\lim_{x\to 0} \frac{\ln(e^x-2x)}{x} $$ but what is after this?

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  • $\begingroup$ You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator... $\endgroup$ – the_candyman Dec 9 '18 at 18:52
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Now since the limit is of $0/0$ form we can apply L'Hopital's rule. So differentiate numerator and denominator, we get

$\lim_{x \to 0}(e^x-2)/(e^x-2x)=-1$.

Now the real limit comes out to be $e^{-1}$.

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    $\begingroup$ I need to take a break my God $\endgroup$ – ovil101 Dec 9 '18 at 19:00
  • $\begingroup$ Don't worry it happens. :) $\endgroup$ – Avanish Singh Dec 9 '18 at 19:01
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$$\lim_{x\to 0} (e^x-2x)^{1/x}\neq \lim_{x\to 0} \frac{1}{x}\ln(e^x-2x)$$

$$\lim_{x\to 0} (e^x-2x)^{1/x}=\lim_{x\to 0} e^{\ln{(e^x-2x)}/x}=$$

$$=\exp\left(\lim_{x\to 0}\frac{\ln (e^x-2x)}x\right)$$ Now we can apply L'Hospital Rule, which means differentiating both numerator and denominator $$\exp\left(\lim_{x\to 0}\frac{e^x-2}{e^x-2x}\right)=e^{-1}=\frac1e$$

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For the form of limit 1^(infinity), lim f(x)^g(x) = e^ lim [{f(x) - 1}•g(x)] Using Maclaurin expansion, you will get e^(-x/x) which is e^-1

e^x = 1 + x/1! + (x^2)/2! + (x^3)/3! ...

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HINT

We have that only by standard limits

$$\lim_{x\to 0} (e^x-2x)^\frac{1}{x}=\lim_{x\to 0} e\left(1-\frac{2x}{e^x}\right)^\frac{1}{x}$$

and by $y=\frac1x \to \infty$

$$\left(1-\frac{2x}{e^x}\right)^\frac{1}{x}=\left(1-\frac{2}{ye^{1/y}}\right)^y=\left[\left(1-\frac{2}{ye^{1/y}}\right)^{\frac{ye^{1/y}}2}\right]^{\frac{2}{e^{1/y}}}$$

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