May 10th, 2019, 12:53 AM  #1 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  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: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 
$\displaystyle \lim_{x\to0}\ln\left[(1+x)^{1/x}\right]=\ln e=1$

May 10th, 2019, 02:33 AM  #3 
Senior Member Joined: Oct 2009 Posts: 782 Thanks: 280 
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  
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  Quote:
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  
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  Quote:
 
May 10th, 2019, 07:16 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,923 Thanks: 1518  
May 10th, 2019, 08:09 PM  #7 
Senior Member Joined: Dec 2015 From: somewhere Posts: 510 Thanks: 79 
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  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  Quote:
$$\lim_{x \rightarrow 0} e^x = \lim_{x \rightarrow 0} (x + 1)$$ $$\implies e = \lim_{x \rightarrow 0} (x+1)^{1/x}$$ 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  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,654 Thanks: 2632 Math Focus: Mainly analysis and algebra  Quote:
\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 lefthanded 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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,654 Thanks: 2632 Math Focus: Mainly analysis and algebra  

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differentiation, limit, logarithm, natural, proove, prove, series, taylor 
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