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