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$$\frac{\ln(e^x+x)}{x}=\frac{e^x+1}{e^x+x}$$

I see that they did something to get rid of the natural log. I couldn't find any properties that would allow me to do this. I also think that they raised both the numerator and denominator by $e$. I have tried it and I did not get the same result. Does anyone know how the solution manual got this? I am supposed to use L'Hopital's rule to find the limit as x approaches $0$.

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    $\begingroup$ What is your question? $\endgroup$ – Pedro Tamaroff May 24 '13 at 1:26
  • $\begingroup$ They are not equivalent expressions. However, their limits as $x$ approaches $0$ are equivalent by L'Hospital's rule. Differentiate the numerator and denominator to see this. $\endgroup$ – Alex Wertheim May 24 '13 at 1:29
  • $\begingroup$ Those two aren't equal; that's actually the application of L'Hopital's rule (differentiating the numerator and denominator). $\endgroup$ – Andy Bromberg May 24 '13 at 1:29
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$$\frac{\ln(e^x+x)}{x} \; \neq \;\;\frac{e^x+1}{e^x+x}$$

It's the application of l'Hospital's rule to the left hand side: take the derivative of the LHS's numerator and of the denominator.

$$\large \frac{e^x+1}{e^x+x}=\frac{\frac{e^x+1}{e^x+x}}{1} = \frac{\frac d{dx}\left({\ln(e^x+x)}\right)}{\frac d{dx}(x)}$$

So $$\lim_{x\to 0} \frac{\ln(e^x+x)}{x} = \lim_{x\to 0} \frac{e^x+1}{e^x+x}$$

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  • $\begingroup$ I have applied L'Hopital's rule but I did not get $e^x+1$ in the numerator, I only got $e^x$. Did I forget to do something? $\endgroup$ – Kot May 24 '13 at 1:29
  • $\begingroup$ $\frac{d}{dx}(e^{x} + x) = e^{x} + 1$ $\endgroup$ – Alex Wertheim May 24 '13 at 1:29
  • $\begingroup$ Oops, I see my mistake. I forgot to apply chain rule since $x$ can be differentiated. $\endgroup$ – Kot May 24 '13 at 1:30
  • $\begingroup$ Exactly...chain rule will get you what you need. $\endgroup$ – Namaste May 24 '13 at 1:38
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    $\begingroup$ @amWhy: $cos(0)$ :-) $\endgroup$ – Amzoti May 24 '13 at 2:33
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The equation that you wrote is false. The correct statement is an application of l’Hospital’s rule:

$$\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{e^x+1}{e^x+x}\;.$$

Since both $\ln(e^x+x)$ and $x$ approach $0$ as $x\to 0$, you can apply l’Hospital’s rule to evaluate the limit. To do this you must differentiate the numerator and the denominator separately.

$$\frac{d}{dx}\ln(e^x+x)=\frac{\frac{d}{dx}(e^x+x)}{e^x+x}=\frac{e^x+1}{e^x+x}\;,$$

and of course $\frac{dx}{dx}=1$, so

$$\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}\;.$$

It is not true without the limits.

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