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May 10th, 2019, 12:53 AM   #1
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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
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May 10th, 2019, 02:15 AM   #2
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$\displaystyle \lim_{x\to0}\ln\left[(1+x)^{1/x}\right]=\ln e=1$
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May 10th, 2019, 02:33 AM   #3
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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?
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May 10th, 2019, 02:53 AM   #4
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle \lim_{x\to0}\ln\left[(1+x)^{1/x}\right]=\ln e=1$
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.
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May 10th, 2019, 02:57 AM   #5
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Quote:
Originally Posted by Micrm@ss View Post
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?
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May 10th, 2019, 07:16 AM   #6
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Originally Posted by alex77 View Post
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 ...

$\dfrac{\ln(x+1)}{x} = \dfrac{1}{x} \cdot \ln(x+1) = \ln(x+1)^{1/x}$
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May 10th, 2019, 08:09 PM   #7
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When x tends to 0 then $\displaystyle e^x \approx x+1$ or $\displaystyle e=\lim_{x\rightarrow 0} (x+1)^{1/x}$.
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May 13th, 2019, 06:40 PM   #8
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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 $\lim_{x \rightarrow 0} (x+1)^{1/x} = \lim_{x \rightarrow 0} 1^{1/x} = e$.
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May 14th, 2019, 07:03 AM   #9
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Quote:
Originally Posted by Micrm@ss View Post
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?
Definitely. But, if you have the integral definition of the logarithm (in which case, why wouldn't you have differentiation available?), that is $$\newcommand{\dt}{\,\mathrm d t}\newcommand{\e}{\mathrm e}\log x = \int_1^x \tfrac1t \dt$$then for some positive real number \(x\) we can consider $t \in \mathbb R$ such that
\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).
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May 14th, 2019, 08:13 AM   #10
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Originally Posted by v8archie View Post
(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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