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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.

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  1. anonymous
    • one year ago
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    Familiar, isn't it? @IrishBoy123

  2. IrishBoy123
    • one year ago
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    indeed! 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.

  3. IrishBoy123
    • one year ago
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    \[\lim_{x\to0}\frac{\ln(\tan x)-\ln x}{e^{x^2}-1}\] \[=\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}-1}\] using Taylor expansions around x=0, for tan x, log x and e^x functions: \(\tan x = x+\frac{ x^3}{3}+\frac{2x^5}{15}-... \) \(\frac{tan \ x}{x} = \frac{x+\frac{x^3}{3}+\frac{2x^5}{15}-...}{x} = 1 + \frac{x^2}{3} + O(x^4)\) \( ln (\frac{\tan x}{ x}) = ln (1 + \frac{x^2}{3} + O(x^4)) \) \(ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...\) \(ln (1 + \frac{x^2}{3} ) = \frac{x^2}{3} - \frac{x^4}{18} + O(x^4)\) \(e^x = 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + ...\) \(e^{x^2} = 1 + x^2+ \frac{x^4}{2!} + \frac{x^6}{3!} + ...\) \( e^{x^2} - 1 = x^2+ O(x^4)\) \[\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}-1}\] \[= \lim_{x\to0} \frac{\frac{x^2}{3} - \frac{x^4}{18} + O(x^4)}{ x^2+ O(x^4)}\] \[= \lim_{x\to0} \frac{\frac{1}{3} - \frac{x^2}{18} + O(x^2)}{ 1+ O(x^2)}\] \[=\frac{1}{3}\]

  4. thomas5267
    • one year ago
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    What are the conditions for the limit of the Taylor series equal to the limit itself?

  5. Empty
    • one year ago
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    Wait! 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!

  6. IrishBoy123
    • one year ago
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    the proof of L'Hopital that i recall uses Maclaurin, so this all sounds very circular. begs the question, what is "the rule"?

  7. IrishBoy123
    • one year ago
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    maybe 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 ?!

  8. IrishBoy123
    • one year ago
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    ouch, that looks like MAclaurin and makes your point but i can see it geometrically too |dw:1442128314117:dw|

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