# Natural logs with L'Hopital's rule

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?

• You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator... – the_candyman Dec 9 '18 at 18:52

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 $$e^{-1}$$.

• I need to take a break my God – ovil101 Dec 9 '18 at 19:00
• Don't worry it happens. :) – Avanish Singh Dec 9 '18 at 19:01

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

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

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

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! ...

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 $$y=\frac1x \to \infty$$

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