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EpsilonDelta Limit Proof 
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#1
Aug1206, 06:35 PM

P: 12

Can anyone prove that the limit of e^x as x > 1 is e using the epsilondelta method?
This is not a homework problem, but I am trying to review my course in analysis from a few years back, and this one has me stumped. 


#2
Aug1206, 08:12 PM

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PF Gold
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Exactly how are you defining e^{x}? One standard way is to define a^{n}, for n a positive integer, as [itex]a\ddot a\cdot\cdot\cdot a[/itex] n times. Then a^{0}= 1 follows from a^{n+ m}= a^{n}a^{m}, a^{n}= 1/a^{n} follows from the same, [itex]a^{\frac{1}{n}}= ^n/sqrt{a}[/itex], [itex]a^{\frac{m}{n}}= ^n\sqrt{a^m}[/itex] which give a^{r} for r any rational number. To define a^{x} for x an arbitrary real number, we take a sequence of rational numbers {r_{n}} that converges to x and define [itex]a^x= \lim_{n\leftarrow \infty}a^{r_n}[/itex]. Of course, that defines a^{x} to be continuous for all x so e^{x} goes to e as x goes to 1 by continuity.
Another method is to define [itex]ln(x)= \int_0^x \frac{1}{x}dx[/itex] and then define e[sup]x[sup] to be the inverse function to ln(x). Of course, since ln(x) is defined as an integral, it is continuous and so is its inverse. That is, e^{x} is continuous by definition and again e^{x} goes to e as x goes to 1 by continuity. 


#3
Aug1206, 08:57 PM

P: 12

I'm simply using e = 2.718... and raising it to x, nothing fancy about it. I'm looking for the trick that will give me delta in terms of epsilon (the epsilondelta proof) or somehow to put a bound on delta.
Unfortunately, for this, I'm not at liberty to make definitions. It's not merely proving the limit, but utilizing the epsilondelta approach. Thanks, though. (I'm preparing for an entrance exam.) 


#4
Aug1306, 12:16 AM

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EpsilonDelta Limit Proof
what are the rest of your decimals? unlkess you tell us exactly what number e is, you cannot prove that e^x really approaches e as x goes to 1 by your method. and unless you define precisely what e^x means you also cannot do it.
the easiest way is halls second method that e^x is continuous, since it is the inverse of the continuous function ln(x), hence it suffices to show e^1 = e, but that follows from the fact that e is defined as the unique number such that ln(e) = 1. 


#5
Aug1306, 05:08 AM

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PF Gold
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If this is a specific question on a practice exam, please state the complete problem itself. 


#6
Aug1306, 07:36 AM

P: 34

[itex] e^x  e  < \epsilon [/itex] for all [itex]1\delta < x < 1+\delta [/itex] and [itex] x \neq 1[/itex]. We have [tex] e^x  e  < \epsilon \Rightarrow \epsilon < e^x  e < \epsilon \Rightarrow e  \epsilon < e^x < e + \epsilon \Rightarrow \ln (e\epsilon) < x < \ln(e + \epsilon) \Rightarrow 1 \ln \frac{1}{1\epsilon / e} < x < 1 + \ln (1+\epsilon/ e) [/tex] (we have used the fact that exponential an logarithmic functions are increasing in their domain). Now, by noting that [tex]\ln(1 + \epsilon/ e) < \ln \frac{1}{1\epsilon /e}[/tex] we can choose [itex]\delta = \ln (1 + \epsilon / e) [/itex] which satisfies our initial request. 


#7
Aug1306, 08:28 AM

P: 12

Gee, I didn't think that defining e was such a priority. I simply meant that e is the irrational number that serves as the base of the natural log, and as such, it can't be stated as a number with a definite number of decimal places (it is like PI in this regard).
Thanks, WigneRacah, you hit the nail right on the head! 


#8
Aug1306, 08:52 AM

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i agree the argument just given is nice and simple, but it proceeds bya ssuming all the thigns that have been pointed out as necessary. the hard part is to prove that log and exp do have the properties that were just assumed, using suitable definitions.
the argument just given is the trivial observation that if f and g are mutually inverse continuous, hence monotone (e.g. increasing) functions, then f maps the interval (ae,a+e) to the interval (f(a)d1,f(a)+d2) if and only if d1 = f(a)  f(ae), and d2 = f(a+e)  f(a). 


#9
Aug1306, 09:04 AM

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for instance, without a definition of log or of e, yopu have no idea what the number delta is that you have been given here, i.e. how can you say that delta must be ln(1 + epsilon/e) when you say you don't need a definition of e? if you don't know what e is then you don't know what ln(1 + epsilon/e) is either.
do you see? you are sort of asking for an answer that will look ok, on a test but without understanding what any of the letters mean in your answer. this is not your fault, as you seem not to have had any instruction in mathematical thinking or proof. the idea is to ask yourself what the symbols mean in your statements. 


#10
Aug1306, 04:17 PM

P: 998

[tex]e = \lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n,[/tex] for example (you have to prove that the limit exists in order for this to make sense, of course). There are many other possibilities. 


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