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anonymous
 one year ago
Compute the limit:
\[\lim_{x\to0}\frac{\ln(\tan x)\ln x}{e^{x^2}1}\]
Oh, and just for fun: no L'Hopital's rule allowed.
anonymous
 one year ago
Compute the limit: \[\lim_{x\to0}\frac{\ln(\tan x)\ln x}{e^{x^2}1}\] Oh, and just for fun: no L'Hopital's rule allowed.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Familiar, isn't it? @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.6indeed! looks a lot scarier too, but merging the top and working with Maclaurin would be my way in i'll type my attempt in later.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.6\[\lim_{x\to0}\frac{\ln(\tan x)\ln x}{e^{x^2}1}\] \[=\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}1}\] using Taylor expansions around x=0, for tan x, log x and e^x functions: \(\tan x = x+\frac{ x^3}{3}+\frac{2x^5}{15}... \) \(\frac{tan \ x}{x} = \frac{x+\frac{x^3}{3}+\frac{2x^5}{15}...}{x} = 1 + \frac{x^2}{3} + O(x^4)\) \( ln (\frac{\tan x}{ x}) = ln (1 + \frac{x^2}{3} + O(x^4)) \) \(ln(1+x) = x  \frac{x^2}{2} + \frac{x^3}{3} + ...\) \(ln (1 + \frac{x^2}{3} ) = \frac{x^2}{3}  \frac{x^4}{18} + O(x^4)\) \(e^x = 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + ...\) \(e^{x^2} = 1 + x^2+ \frac{x^4}{2!} + \frac{x^6}{3!} + ...\) \( e^{x^2}  1 = x^2+ O(x^4)\) \[\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}1}\] \[= \lim_{x\to0} \frac{\frac{x^2}{3}  \frac{x^4}{18} + O(x^4)}{ x^2+ O(x^4)}\] \[= \lim_{x\to0} \frac{\frac{1}{3}  \frac{x^2}{18} + O(x^2)}{ 1+ O(x^2)}\] \[=\frac{1}{3}\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0What are the conditions for the limit of the Taylor series equal to the limit itself?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Wait! Isn't the Maclauren series the same as differentiating essentially, so I am still curious about how you could do this without using L'Hopital's rule!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.6the proof of L'Hopital that i recall uses Maclaurin, so this all sounds very circular. begs the question, what is "the rule"?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.6maybe i've been carrying the wrong baggage around all this time but i figured that \[\frac{f(x+\delta )}{g(x+\delta )}= \frac{f(x) + \delta f'(x) + ...}{g(x) + \delta g'(x) + ...}\] so f(x) = g(x) = 0 allows you to look at the derivatives meaning l'hopital's rule is just a black box? too hand wavey ?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.6ouch, that looks like MAclaurin and makes your point but i can see it geometrically too dw:1442128314117:dw
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