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I was doing a derivative by definition and I need to solve that limit using equivalent infinitesimals and such, without L'Hopital Rule. Any hint?

FShrike
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user10000024
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    Forget L'Hôpital's rule. This is just the definition of the derivative of $\log(x)$ at $x=e$. "Equivalent infinitesimals"? Rant: Many people use L'Hôpital's rule when they are just using the definition of the derivative; this is highly circular reasoning. – Ted Shifrin Jan 25 '22 at 20:12
  • I do know that is the definition of that derivative, but I cannot use derivatives either. By equivalent infinitesimals I mean, for instance, $x\sim sin(x)$ when $x\to 0$, so we can substitute $x$ by $sin(x)$ in limits. – user10000024 Jan 25 '22 at 20:14
  • Well, you need to make it completely explicit what you know, then. We can't guess. – Ted Shifrin Jan 25 '22 at 20:16
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    Where does the formal justification for $x\sim\sin(x)$ come from? Taylor series and derivatives... it's really all the same – FShrike Jan 25 '22 at 20:18
  • @FShrike Yes, of course, but we assume the OP's teacher has given specific and limiting instructions. But we can't guess what is in the toolbox. – Ted Shifrin Jan 25 '22 at 20:28
  • Can you use that $\lim_{x\to 0}\frac{\log(1+x)}{x}=1$? – Vasili Jan 25 '22 at 20:31
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    As for a hint, (I just deleted my answer because I realised you didn't want a full answer), if you are happy to use $x\sim\sin x$ then you should also be happy using $\log(1+x)\sim x$, which will give you the answer. However, for future questions you really should be clearer on what you're allowed to use and what you know – FShrike Jan 25 '22 at 20:32
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    Does this answer your question? Determine $\lim_{h \to 0}\frac{\ln(e+h)-1}{h}$ – Martin R Jan 25 '22 at 20:50
  • @PedroGarcía Note you can't just replace $\sin x$ by $x$ in any limit where $x$ tends to $0$. See e.g. here https://math.stackexchange.com/questions/2584328/why-can-we-replace-an-infinitesimal-in-a-limit-with-an-equivalent-infinitesimal – Snaw Jan 25 '22 at 20:50

3 Answers3

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Hint: $$\lim_{x\to 0}\frac{\log(e+x)-1}{x}=\lim_{x\to 0}\frac{\log(e+x)-\ln e}{x}=\frac{1}{e}\lim_{x\to 0}\frac{\log(1+\frac{x}{e})}{\frac{x}{e}}$$ and use equivalency $x \sim \ln(1+x)$

Vasili
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$$\lim_{x\to0}\frac{\log e(1+\frac xe) -1}x\\=\lim_{x\to0}\frac{\log e+\log(1+\frac xe) -1}x\\=\lim_{x\to0}\frac {\log(1+\frac xe) }x\\=\lim_{x\to0}\frac{\frac xe-\frac{(\frac xe)^ 2}2+...}x\\=\frac 1e$$

Vasili
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aarbee
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We can define $\ln x=\int_1^x(1/t)dt$ for $x>0.$ With the substitution $u=At$ we have $$\ln A+\ln B=\int_1^A(1/t)dt +\int_1^B(1/u)du=$$ $$=\int_1^A(1/t)dt+\int_{t=A}^{AB}(1/At)dAt=$$ $$=\int_1^A(1/t)dt+\int_A^{AB}(1/t)dt=$$ $$=\int_1^{AB}(1/t)dt=\ln AB.$$

For your Q, we have $-1+\ln (e+x)=-1+\ln e(1+x/e)=$ $=-1+1+\ln (1+x/e)=\ln (1+x/e).$

Observe that if $1+x/e> 0$ then $$\frac {x/e}{1+x/e}=\int_1^{1+x/e}\frac {1}{1+x/e}dt\le $$ $$\le \int_1^{1+x/e}(1/t)dt=\ln (1+x/e)\le $$ $$\le \int_1^{1+x/e}1\cdot dt=x/e.$$

DanielWainfleet
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  • If we define $\ln x$ this way, we can easily show that its inverse is $e^x$ where $e$ is the unique number such that $\int_1^e(1/t)dt=1.$ – DanielWainfleet Jan 25 '22 at 22:16