Why does $(1+frac{1}{n})^n$ as $n toinfty = e$ mean the same thing as $(e^x)'$ = $e^x$












0














I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.



I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.



I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.



By the way, here is the work that I have done so far:



$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$



This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!






algebra-precalculus exponential-function







share|cite|improve this question




















  • 2




    The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
    – 0x539
    Nov 22 '18 at 22:38






  • 1




    A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
    – Yuriy S
    Nov 22 '18 at 22:39










  • math.stackexchange.com/questions/2067849/…
    – amWhy
    Nov 22 '18 at 22:51










  • You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
    – gimusi
    Nov 22 '18 at 22:59










  • @gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
    – amWhy
    Nov 22 '18 at 23:02
















0














I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.



I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.



I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.



By the way, here is the work that I have done so far:



$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$



This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!






algebra-precalculus exponential-function







share|cite|improve this question




















  • 2




    The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
    – 0x539
    Nov 22 '18 at 22:38






  • 1




    A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
    – Yuriy S
    Nov 22 '18 at 22:39










  • math.stackexchange.com/questions/2067849/…
    – amWhy
    Nov 22 '18 at 22:51










  • You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
    – gimusi
    Nov 22 '18 at 22:59










  • @gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
    – amWhy
    Nov 22 '18 at 23:02














0












0








0







I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.



I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.



I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.



By the way, here is the work that I have done so far:



$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$



This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!






algebra-precalculus exponential-function







share|cite|improve this question















I would like a proof as to why as the limit of $n$ approaches infinity, $(1+frac{1}{n})^n$ means the same thing as the derivative of $e^x$ = $e^x$.



I have tried proving this by converting $(e^x)'$ into the limit definition of e, and vice versa but keep ending up at a dead end.



I am a Precal BC student, so I do not have much Calculus experience and would love an explanation that involves as little Calculus as possible.



By the way, here is the work that I have done so far:



$(e^x)'$ = $e^x$ $Rightarrow$ derivative formula = $lim_{h to 0} frac{f(x+h)- f(x)}{h}$ $Rightarrow$ plug it in $Rightarrow$ $lim_{h to 0} frac{e^{(x+h)} - e^x}{h}$ = $e^x Rightarrow lim_{h to 0} frac{e^x(e^h - 1)}{h}$ = $e^x$ $Rightarrow$ h = $frac{1}{n}$ $Rightarrow$ $n to infty$ $Rightarrow$ $lim_{n to infty}$ $frac{e^x(e^{(1/n)} - 1)}{frac{1}{n}}$ = $e^x$ $Rightarrow$ $lim_{n to infty} (e^{frac{1}{n}} - 1)n = 1$



This is the part where I get stuck. I am not supposed to transfer the limit over to the other side so I don't know what to do. Please help and thanks in advance!






algebra-precalculus exponential-function




algebra-precalculus exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 13:33









user21820

38.8k543153




38.8k543153










asked Nov 22 '18 at 22:36









Freedom EagleFreedom Eagle

162




162








  • 2




    The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
    – 0x539
    Nov 22 '18 at 22:38






  • 1




    A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
    – Yuriy S
    Nov 22 '18 at 22:39










  • math.stackexchange.com/questions/2067849/…
    – amWhy
    Nov 22 '18 at 22:51










  • You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
    – gimusi
    Nov 22 '18 at 22:59










  • @gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
    – amWhy
    Nov 22 '18 at 23:02














  • 2




    The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
    – 0x539
    Nov 22 '18 at 22:38






  • 1




    A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
    – Yuriy S
    Nov 22 '18 at 22:39










  • math.stackexchange.com/questions/2067849/…
    – amWhy
    Nov 22 '18 at 22:51










  • You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
    – gimusi
    Nov 22 '18 at 22:59










  • @gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
    – amWhy
    Nov 22 '18 at 23:02








2




2




The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
– 0x539
Nov 22 '18 at 22:38




The limit is simply $e$ (without any derivative or exponent). You should maybe say what your definition of $e$ is since some people would define it exactly that way.
– 0x539
Nov 22 '18 at 22:38




1




1




A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
– Yuriy S
Nov 22 '18 at 22:39




A lot of people use this limit as the definition of $e$. We can prove that it's equivalent to other definitions and find the decimal approximation to the number with it
– Yuriy S
Nov 22 '18 at 22:39












math.stackexchange.com/questions/2067849/…
– amWhy
Nov 22 '18 at 22:51




math.stackexchange.com/questions/2067849/…
– amWhy
Nov 22 '18 at 22:51












You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
– gimusi
Nov 22 '18 at 22:59




You should clarify what exactly you are looking for. Are you aware that $(1+frac{1}{n})^nto e$? Whay are you referring to $e^x$?
– gimusi
Nov 22 '18 at 22:59












@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
– amWhy
Nov 22 '18 at 23:02




@gimusi That sort of clarification ought to be sought, and obtained, before jumping the gun to answer.
– amWhy
Nov 22 '18 at 23:02










1 Answer
1






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0














That limit expression gives the number $e$ itself: for $e^x$ it's



$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$



if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,



$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$



& it would make no essential difference to it, because of $n$ tending to $infty$.






share|cite|improve this answer





















  • How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
    – Freedom Eagle
    Nov 25 '18 at 0:31










  • Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
    – Freedom Eagle
    Nov 25 '18 at 0:52












  • It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
    – AmbretteOrrisey
    Nov 25 '18 at 10:34












  • The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
    – AmbretteOrrisey
    Nov 25 '18 at 10:38













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1 Answer
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1 Answer
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0














That limit expression gives the number $e$ itself: for $e^x$ it's



$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$



if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,



$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$



& it would make no essential difference to it, because of $n$ tending to $infty$.






share|cite|improve this answer





















  • How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
    – Freedom Eagle
    Nov 25 '18 at 0:31










  • Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
    – Freedom Eagle
    Nov 25 '18 at 0:52












  • It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
    – AmbretteOrrisey
    Nov 25 '18 at 10:34












  • The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
    – AmbretteOrrisey
    Nov 25 '18 at 10:38


















0














That limit expression gives the number $e$ itself: for $e^x$ it's



$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$



if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,



$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$



& it would make no essential difference to it, because of $n$ tending to $infty$.






share|cite|improve this answer





















  • How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
    – Freedom Eagle
    Nov 25 '18 at 0:31










  • Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
    – Freedom Eagle
    Nov 25 '18 at 0:52












  • It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
    – AmbretteOrrisey
    Nov 25 '18 at 10:34












  • The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
    – AmbretteOrrisey
    Nov 25 '18 at 10:38
















0












0








0






That limit expression gives the number $e$ itself: for $e^x$ it's



$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$



if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,



$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$



& it would make no essential difference to it, because of $n$ tending to $infty$.






share|cite|improve this answer












That limit expression gives the number $e$ itself: for $e^x$ it's



$$lim_{nrightarrowinfty}left(1+frac{x}{n}right)^n .$$



if you differentiate this, you get a $1/n$ from inside - the coefficient of $x$; and a $n$ from the exponent, which cancel out. You are left with $n-1$ as the exponent ... but this makes no difference in the limit as $nrightarrowinfty$ ... infact you could take that limit formula & add any finite real number to either occurence of $n$ in it,



$$lim_{nrightarrowinfty}left(1+frac{x}{n+mu}right)^{n+nu} ,$$



& it would make no essential difference to it, because of $n$ tending to $infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 '18 at 15:43









AmbretteOrriseyAmbretteOrrisey

57410




57410












  • How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
    – Freedom Eagle
    Nov 25 '18 at 0:31










  • Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
    – Freedom Eagle
    Nov 25 '18 at 0:52












  • It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
    – AmbretteOrrisey
    Nov 25 '18 at 10:34












  • The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
    – AmbretteOrrisey
    Nov 25 '18 at 10:38




















  • How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
    – Freedom Eagle
    Nov 25 '18 at 0:31










  • Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
    – Freedom Eagle
    Nov 25 '18 at 0:52












  • It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
    – AmbretteOrrisey
    Nov 25 '18 at 10:34












  • The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
    – AmbretteOrrisey
    Nov 25 '18 at 10:38


















How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
– Freedom Eagle
Nov 25 '18 at 0:31




How would the expression of $e^x$ be lim n→∞ $(1+/frac{x}{n})^n$ rather than it being $(1+/frac{1}{n})^xn$? And, by the way, how did you get your limit sign to look like that?
– Freedom Eagle
Nov 25 '18 at 0:31












Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
– Freedom Eagle
Nov 25 '18 at 0:52






Really sorry for the formatting issues. What I meant to ask was how the expression of $e^x$ be lim n→∞ $(1+frac{x}{n})^n$ rather than it being $(1+frac{1}{n})^{xn}$?
– Freedom Eagle
Nov 25 '18 at 0:52














It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
– AmbretteOrrisey
Nov 25 '18 at 10:34






It pans out equivalent - precisely because of the limiting process. Observe how it's equivalent to $$lim_{frac{n}{x}rightarrowinfty}left(1+frac{x}{n}right)^{xfrac{n}{x}}$$
– AmbretteOrrisey
Nov 25 '18 at 10:34














The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
– AmbretteOrrisey
Nov 25 '18 at 10:38






The letting $nrightarrowinfty$ allows various strangelooking 'liberties' to be taken ... like that substitution of $mu$ & $nu$ that I did. And you'll see I've answered your other question!
– AmbretteOrrisey
Nov 25 '18 at 10:38




















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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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