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
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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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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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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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