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 May 10th, 2019, 12:53 AM #1 Member   Joined: Aug 2016 From: Romania Posts: 30 Thanks: 1 Prove the limit of natural logarithm without differentiation or Taylor series Can someone help me to prove that this limit is equal to 1 without using differentiation or Taylor Series?(like in the limit of sinx/x when x tends to 0)What are the steps? https://imgur.com/a/WslKnLI
 May 10th, 2019, 02:15 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $\displaystyle \lim_{x\to0}\ln\left[(1+x)^{1/x}\right]=\ln e=1$ Thanks from Maschke
 May 10th, 2019, 02:33 AM #3 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 327 Dear alex, The answer of greg is undoubtedly correct, but is most likely not the answer you are looking for. These questions often pop up when trying to establish basic facts. So the above proof might not work becuase certain facts have not yet been established. In this case, you need to tell us quite preciely what you're working with. What are the definitions of e, ln, exponentiation. What properties have you already proven, what can be used, etc?
May 10th, 2019, 02:53 AM   #4
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 Originally Posted by greg1313 $\displaystyle \lim_{x\to0}\ln\left[(1+x)^{1/x}\right]=\ln e=1$
How is 1/x raised to the power of ln(x+1)? I don't understand,please explain.
I know that 1/x is not the exponenent of ln(x+1).
Please explain how ln(x+1) got the exponent 1/x when 1/x is the denominator.

May 10th, 2019, 02:57 AM   #5
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 Originally Posted by Micrm@ss Dear alex, The answer of greg is undoubtedly correct, but is most likely not the answer you are looking for. These questions often pop up when trying to establish basic facts. So the above proof might not work becuase certain facts have not yet been established. In this case, you need to tell us quite preciely what you're working with. What are the definitions of e, ln, exponentiation. What properties have you already proven, what can be used, etc?
I'm trying to prove that ln(x+1)/x is equal to 1 but I don't know what steps to use.Suppose that I use that proof of ln(1+x)^1/x,how is this proof going to help me understand that lne=1?

May 10th, 2019, 07:16 AM   #6
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 Originally Posted by alex77 Please explain how ln(x+1) got the exponent 1/x when 1/x is the denominator.
correction ... $x$ is the denominator

note the power property of logarithms ...

$a \log{b} = \log{b^a}$

ergo ...

$\dfrac{\ln(x+1)}{x} = \dfrac{1}{x} \cdot \ln(x+1) = \ln(x+1)^{1/x}$

 May 10th, 2019, 08:09 PM #7 Senior Member   Joined: Dec 2015 From: somewhere Posts: 606 Thanks: 88 When x tends to 0 then $\displaystyle e^x \approx x+1$ or $\displaystyle e=\lim_{x\rightarrow 0} (x+1)^{1/x}$.
May 13th, 2019, 06:40 PM   #8
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 Originally Posted by idontknow When x tends to 0 then $\displaystyle e^x \approx x+1$ or $\displaystyle e=\lim_{x\rightarrow 0} (x+1)^{1/x}$.
Wait a minute... So we have:

$$\lim_{x \rightarrow 0} e^x = \lim_{x \rightarrow 0} (x + 1)$$
$$\implies e = \lim_{x \rightarrow 0} (x+1)^{1/x}$$

By that logic, we could replace $e$ with anything and the proof would still work. E.g.:

$$\lim_{x \rightarrow 0} 9^x = \lim_{x \rightarrow 0} (x + 1)$$
$$\implies 9 = \lim_{x \rightarrow 0} (x+1)^{1/x}$$

While we're at it, since $\lim_{x \rightarrow 0} (x+1) = 1$, we could also conclude that $\lim_{x \rightarrow 0} (x+1)^{1/x} = \lim_{x \rightarrow 0} 1^{1/x} = e$.

May 14th, 2019, 07:03 AM   #9
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 Originally Posted by Micrm@ss These questions often pop up when trying to establish basic facts. So the above proof might not work becuase certain facts have not yet been established. In this case, you need to tell us quite preciely what you're working with. What are the definitions of e, ln, exponentiation. What properties have you already proven, what can be used, etc?
\begin{align*}1 &\lt t \lt 1 + \frac1x \\ \implies \frac{1}{1+\frac1x} &\lt \frac1t \lt 1\end{align*}
This inequality is, by definition, valid for all $t$ in the open interval $(1,1+\frac1x)$,
so we can integrate over that region and the inequality will continue to hold. \begin{align*}
\int_1^{1+\frac1x} \frac{1}{1+\frac1x} \dt &\lt \int_1^{1+\frac1x} \frac1t \dt \lt \int_1^{1+\frac1x} \dt \\
\frac1x \cdot \frac{1}{1+\frac1x} = \frac{1}{x+1} &\lt \log{\left( 1+\frac1x \right)} \lt \frac1x \\
{x \over x+1} &\lt x\log{\left( 1+\frac1x \right)} \lt 1 \\
{x \over x+1} &\lt \log{\left(\left( 1+\frac1x \right)^x\right)} \lt 1 \\
\end{align*}
Now take the limit as $$x \to +\infty$$.
Since the outer two limits are both 1, the squeeze theorem gives us
$$\lim_{x \to +\infty} \log{\left(\left( 1+\frac1x \right)^x\right)} = 1$$
Writing $y = \frac1x$ then give us $$\lim_{y \to 0^+} \log{\left(\left( 1+y \right)^{\frac1y}\right)} = 1$$
I'll leave the left-handed limit for someone else to consider (it's straightforward starting from the line above where we took the limit to $+\infty$ before).

May 14th, 2019, 08:13 AM   #10
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 Originally Posted by v8archie (it's straightforward starting from the line above where we took the limit to $+\infty$ before)
Actually, it's a little more complicated than that since we originally defined $x$ to be positive.

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