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March 12th, 2017, 08:33 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
How do you know the Taylor series for $\cosh x$? $\dfrac{e^x + e^{x} 2}{x^2} = \left(\dfrac{\sinh u}{u}\right)^2$, where $u = \dfrac{x}{2}$. 
March 12th, 2017, 08:50 PM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra  
March 13th, 2017, 01:24 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
Its usual proof assumes (amongst other things) that its derivative is $e^x$. By definition, that derivative is $\displaystyle \lim_{h\to0}\frac{e^{x + h}  e^x}{h} = e^x \lim_{h\to0}\frac{e^h  1}{h}$. Thus, its proof rests on knowing a limit that is very similar to (in fact, equivalent to) the one you're trying to evaluate. 
March 13th, 2017, 03:03 AM  #14  
Senior Member Joined: Sep 2016 From: USA Posts: 356 Thanks: 194 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x}$$ which many would claim can be computed using L'Hospital. However, using L'Hospital requires knowing the derivative of $\sin x$ exists and you may recognize that the limit we are computing is precisely this derivative at $x=0$. In fact, if you apply L'Hospital to the given example, this will be exactly the logical flaw you run into in the 2nd application. The limit you are computing is actually the definition of the derivative of a function you must assume the derivative of in order to apply it. It is completely circular.  
March 13th, 2017, 04:18 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
The difference being that it isn't necessary to know the limit to compute the derivative. For example, if your definition is that $f(x)=e^x$ is the inverse of the function $\ln{(x)} = \int_{1}^x \frac1t \,\mathrm dt$, we get the derivative via $$\begin{align*} y &= f(x) &\implies g(y) &= x \\ \implies g'(y)y' &= 1 &\implies y' = f'(x) &= \frac1{g'(y)} = \frac1{g'\big(f(x)\big)} \end{align*}$$ Or, if one defines $f(x)=e^x$ as the function that is it's own derivative,we get the limit from the derivative and not vice versa. Even if your definition is that $$f(x)=e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n$$ The derivative comes without knowledge of the limit (although in this case we don't need either the derivative or the limit for this problem). Last edited by v8archie; March 13th, 2017 at 04:51 AM. 
March 15th, 2017, 06:38 AM  #16 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,783 Thanks: 1025 Math Focus: Elementary mathematics and beyond  

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