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Original question is: $$\lim_{x \to 0}\frac {1}{x} \cdot \ln (e^x+x)$$ Which is $$\frac 00$$ So I use L'Hôpital's rule and use the derivative but I'm not sure how to do that. I tried and got: $$-\frac {1}{x^2}\cdot\frac{1}{e^x+x}$$ which doesnt help me at all... Thanks

Nathan Maschler
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4 Answers4

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l'Hopital's rule says that if $f\to0$ and $g\to0$ (at the same point $a$) and both functions are derivable at this point, then $$\lim_{x\to a}\frac fg=\lim_{x\to a}\frac{f'}{g'}$$

Your limit can be written this way:

$$\lim_{x\to0}\frac{\ln(e^x+x)}{x}$$

Since numerator and denominator tend to $0$, this is the same as $$\lim_{x\to0}\frac{\frac{e^x+1}{e^x+x}}{1}=2$$

ajotatxe
  • 65,084
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Hint: using the chain rule, $$ \frac{d}{dx} \ln ( e^x + x ) = \frac{\frac{d}{dx} (e^x + x)}{e^x + x} = \frac{e^x + 1}{e^x + x}. $$

Sammy Black
  • 25,273
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Using L'Hôpital's rule $$ \lim\limits_{x\to 0} \frac{\ln(e^x+x)}{x} = \lim\limits_{x\to 0} \frac{\frac{\mathrm d}{\mathrm dx}\left[\ln(e^x+x)\right]}{\frac{\mathrm d}{\mathrm dx}[x]}= \lim\limits_{x\to 0} \frac{\frac{1}{e^x+x}\frac{\mathrm d}{\mathrm dx}\left[e^x+x\right]}{1}= \lim\limits_{x\to 0} \frac{e^x+1}{e^x+x}=\frac{1+1}{1+0}=2 $$ Using equivalent infinitesimals $$\lim\limits_{x\to 0} \frac{\ln(e^x+x)}{x}=\lim\limits_{x\to 0} \frac{e^x+x-1}{x}=\lim\limits_{x\to 0} \left[1+\frac{e^x-1}{x}\right]=1+\lim\limits_{x\to 0} \frac{x\ln e}{x}=2$$

k170
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$$\lim_{x \to 0}\frac{\ln (e^x+x)}{x}\underbrace{=}_{\mathrm{L'Hospital's rule}}\lim_{x\to 0} \frac{\displaystyle \frac{d}{dx} \frac{e^x+1}{e^x+x}}{\displaystyle \frac{d}{dx} x}=\lim_{x\to 0} \frac{\displaystyle \frac{e^x+1}{e^x+x}}{1}=2.$$

mfl
  • 29,399
  • I see what i was doing wrong, thank you. One more question though, why is $$ln(e^x+x)=\frac{e^x+1}{e^x+x}$$ I thought when you get the derivative of $$ln(x)$$ it turns into $$\frac 1x$$ – Nathan Maschler Nov 05 '14 at 19:07
  • Yes, but the derivative of a composite function as $\ln f(x)$ is $\frac{f'(x)}{f(x)}.$ – mfl Nov 05 '14 at 19:53