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My attempt:

$$\lim_{x\to a}\; \frac{e^x - e^a}{x-a} = $$ $$\frac{e^a - e^a}{a-a} = $$ $$\frac{e^a(1 - 1)}{a(1 - 1)} = $$ $$\frac{e^a}{a}$$

My textbook says the correct answer is $e^a$. How do I get rid of the $a$ in my denominator?

Edit:

I don't know anything about calculus. According to my textbook it's possible with just algebra.

user1534664
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    You divided by zero from the second line on... – AlexR Jan 19 '15 at 22:55
  • Multiply by the conjugate – Julian Rachman Jan 19 '15 at 22:56
  • Your second line is $0/0$, so you have an indeterminate form. Do you know anything about derivatives? – user208259 Jan 19 '15 at 22:57
  • possible duplicate of Derivative of exp with definition of differentiability – AlexR Jan 19 '15 at 22:57
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    This is the definition of the derivative of $e^x$ for $x = a$. – Ivo Terek Jan 19 '15 at 22:57
  • I don't know anything about calculus. According to my textbook it's possible with just algebra. – user1534664 Jan 19 '15 at 22:58
  • @user208259 I suspect in a proof that $(e^x)' = e^x$ you can't use l'Hospital on the same function ;) – AlexR Jan 19 '15 at 22:58
  • @AlexR That question involves power series of a complex variable. This one is much more elementary. – user208259 Jan 19 '15 at 22:58
  • @AlexR Who mentioned L'Hopital's Rule? – user208259 Jan 19 '15 at 22:59
  • @user208259 I'm eager to see your proof without using the derivative of $\exp$. Note that the question is a difference quotient, usually introduced as the definition of the derivative. – AlexR Jan 19 '15 at 22:59
  • @AlexR Most likely, this student doesn't have a fully rigorous definition of the exponential function. – user208259 Jan 19 '15 at 23:00
  • @user1534664 Are you aware of any limit relations concerning the number $e$ that have been discussed in your textbook? – user208259 Jan 19 '15 at 23:01
  • @user208259 Be aware that the answer in the duplicate also works in the reals. If the OP doesn't provide his definition of $\exp$, it's fair to assume the power series just as any other definition. – AlexR Jan 19 '15 at 23:04
  • @AlexR We are dealing with a lower level of sophistication. Have a look at Stewart's calculus textbook for example. We can't fault someone for working with what may be a non-rigorous textbook, as about 99% of calculus students do. – user208259 Jan 19 '15 at 23:05
  • I'm unsure what you mean, @user208259. My textbook is terrible, but my school forces it upon me. I think I'll skip this assignment, with all the commotion around here. – user1534664 Jan 19 '15 at 23:09
  • @user1534664 I'm not saying you shouldn't be asking this question. Could you please say what your textbook tells you about the number $e$? How is $e$ defined? – user208259 Jan 19 '15 at 23:11
  • @user208259 The book says e is an irrational number that is used in calculus alot. It also says that all exponential functions have the form $r^x$ and all of them have a tangent line ($y = mx + b$) when $x = 0$. It then says that there's only one value of $r$ when $m = 1$. That value is $e$. After that it starts using it throughout the book. – user1534664 Jan 19 '15 at 23:19
  • Okay, then I imagine that it mentions the result that Ivo Terek assumes you've seen. Is that right? – user208259 Jan 19 '15 at 23:21
  • @user208259 As rediculous as it is, the book shows no intermediate steps. It only shows the question and the answer. So I first try the assignment and then I can check if my answer is correct. I think there's a good chance Ivo's answer is what I need. I will accept it. Thanks. – user1534664 Jan 19 '15 at 23:25
  • @user1534664 If that limit relation isn't mentioned, you can get it by writing down the slope of the tangent line to $e^x$ at $(0,1)$ as a limit. According to your book's definition of $e$, this slope must be $1$. – user208259 Jan 19 '15 at 23:27

1 Answers1

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I suppose that you have already seen the result $$\lim_{x \to 0}\frac{e^x-1}{x} = 1,$$ I'm going to use it. In your limit, call $h = x - a$. So: $$\lim_{x \to a}\frac{e^x-e^a}{x-a}=\lim_{h \to 0}\frac{e^{a+h}-e^a}{h} = \lim_{h \to 0}e^a\,\frac{e^h-1}{h} = e^a.$$

If you use as definition that $$e^x:=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots,$$ then the first result is justified by $$\lim_{x \to 0}\frac{e^x-1}{x} = \lim_{x \to 0}\frac{1+x+{\rm o}(x^2)-1}{x}=\lim_{x \to 0} \,1 + \frac{{\rm o}(x^2)}{x} = 1.$$

Ivo Terek
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  • Where do you justify the essential part $\lim_{h\to0} \frac{e^h - 1}h = 1$? It's still indeterminate. – AlexR Jan 19 '15 at 23:01
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    @AlexR Some textbooks take this as the definition of the number $e$. Without proving its existence or uniqueness, of course. – user208259 Jan 19 '15 at 23:04
  • @user208259 References? I've only come across one of $\sum \frac1{k!}$, $\lim (1+\frac1n)^n$ and slight variants. – AlexR Jan 19 '15 at 23:05
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    I added a little justificative, just to don't let it floating, then. – Ivo Terek Jan 19 '15 at 23:08
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    @IvoTerek Nice now the answer is complete as far as I'm concerned :) (+1) – AlexR Jan 19 '15 at 23:12
  • @AlexR Please see Stewart's Early Transcendentals, 7th Edition, page 180. I'm not defending this on mathematical grounds, but it's something that isn't uncommon. – user208259 Jan 19 '15 at 23:12
  • @user208259 Thanks I was curious, no offence intended and no dispute whatsoever. I'll take a look at it. EDIT: I must admit that I find it confusing because real powers are defined before definition of $\exp$, the usual way to compute them... ($a^b = \exp(b \ln a)$) – AlexR Jan 19 '15 at 23:13
  • @AlexR In most calculus textbooks, people are assumed to know what $a^b$ means when $b$ is rational. Then, at best, it is mentioned without proof somewhere that there is only one way to extend the function $a^x$ to all values of $x$ so that it is continuous (or monotonic). At worst, nothing is mentioned about irrational exponents, because the majority of students don't realize that there's anything to worry about. – user208259 Jan 19 '15 at 23:25
  • @user208259 That makes me sad... I must feel lucky that we didn't have such a professor :D The first thing I typed into my calculator way back was $e^\pi$ just to see how it would do. – AlexR Jan 19 '15 at 23:27
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    @AlexR For an alternative proof of differentiability, see http://math.stackexchange.com/questions/1015202/prove-that-the-exponential-function-is-differentiable In this case, the existence of the exponential function would first need to be proved by either: (1) first defining it for any rational $x$ with a denominator of the form $2^n$ and then using the completeness property of the real numbers, or (2) by using the fact that any two complete totally ordered groups are isomorphic. – user208259 Jan 19 '15 at 23:33