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I want to calculate the limit $\lim_{x\to0} \frac{\ln(x+e)-1}{x}$.

Just putting in the $0$ we can see $\frac{\ln(e)-1}{0} = \frac{0}{0}$ so my first guess is to use "L'Hopital" rules.

$\lim_{x\to0} \frac{\ln(x+e)-1}{x} = \lim_{x\to0} \frac{\frac{1}{x+e}}{1} = \frac{1}{e}$.

edit: I think this is the correct solution, can anyone approve?

Alessio K
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D idsea J
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    Substituting $x=0$ into the original equation gives $\frac{1}{0}$. Check your working again. – Tham Sep 23 '21 at 15:50
  • Oh sorry, I've forgot an important part, will edit it now. I made the same mistake on the graph aswell, I actually got the correct solution if I'm not wrong. – D idsea J Sep 23 '21 at 15:51
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    It looks fine now but it can be done without using L'Hospital. You may recognize that $\ln(x+e)-\ln(e)=\ln(\frac{x}{e}+1)$ – Vasili Sep 23 '21 at 15:55
  • About your original non-edited question, the graph is referred to logarithm with base $10$; instead, you're working with $\text{ln}$ which is in base $e$. Your work is correct, another way to evaluate this is notice that $\ln(x+e)=\ln[e(1+x/e)]=\ln e + \ln(1+x/e)$. – Bernkastel Sep 23 '21 at 15:58

3 Answers3

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Your solution is fine, we can also proceed without l’Hospital as follows

$$\frac{\ln(x+e)-1}{x}= \frac{\ln e+\ln(1+x/e)-1}{x}= \frac1e\frac{\ln(1+x/e)}{x/e}$$

and then conclude by standard limits.

user
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Your work is correct now.

Notice that

$$\frac{\ln(x+e)-1}{x}=\frac{\ln(x+e)-\ln(e)}{x}=\frac{\ln\left(1+\frac{x}{e}\right)}{x}$$

so setting $x=ue$ we have

$$\lim_{x\rightarrow 0}\frac{\ln(x+e)-1}{x}=\frac{1}{e}\lim_{u\rightarrow 0}\frac{\ln(1+u)}{u}=\frac{1}{e}$$

See also here.

Alessio K
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  • Doesn't this only prove that if there is a limit, then it must be $\frac{1}{e}$? It could be that taking the limit along some other path approaching 0 is different. – JLinsta Sep 23 '21 at 16:09
  • No, the last limit is a standard limit. See the link. So the limit exists and it is $\frac{1}{e}$. If $x=ue$, then $x\rightarrow 0$ means that $u\rightarrow 0$ since $e$ is a constant. – Alessio K Sep 23 '21 at 16:10
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This works. If you're interested, an alternative way to attack this problem is to recognize that the limit

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

is the derivative of $\ln$ evaluated at $e$. You know that $\frac{d}{dx}\ln(x)=\frac{1}{x}$ for every positive $x$, so

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

Alann Rosas
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