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SSmartyPants replied to the thread High School Taking a limit and getting the wrong answer...don't know why.Yes, I went ahead and calculated the limit using Taylor series the other night...it's alot easier this way than trying to simplify...
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SSmartyPants reacted to PeroK's post in the thread High School Taking a limit and getting the wrong answer...don't know why with
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Here's a solution using Taylor Series: $$\tan x = x + \frac{x^3}{3} + \dots$$$$\sin x = x - \frac{x^3}{6} + \dots$$$$\tan x - \sin x =... -
SSmartyPants reacted to PeroK's post in the thread High School Taking a limit and getting the wrong answer...don't know why with
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Here's something interesting, and possibly useful: In this case, we needed ##t'''(0)##, where ##t(x) = \tan x##. Note that: $$t'(x) =... -
SSmartyPants reacted to PeroK's post in the thread High School Taking a limit and getting the wrong answer...don't know why with
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You can apply L'Hopital directly. Let ##t(x) = \tan x## and ##s(x) = \sin x##, then: $$t'(x) = \frac 1 {\cos^2 x}, \ t''(x) =... -
SSmartyPants reacted to renormalize's post in the thread High School Taking a limit and getting the wrong answer...don't know why with
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Your step:$$\lim_{x\to0}\frac{\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin... -
SSmartyPants reacted to PeroK's post in the thread High School Taking a limit and getting the wrong answer...don't know why with
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I would have been tempted to try letting ##y = \sin x##. I don't know if that helps. -
SSmartyPants replied to the thread High School Taking a limit and getting the wrong answer...don't know why.I did see an example of this limit being solved with the substitution you mentioned, and it does make things a bit easier to visualize.
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SSmartyPants replied to the thread High School Taking a limit and getting the wrong answer...don't know why.Ah, I see. So alternatively, the limit of the product is NOT the product of the limits when either ##f(x)## or...
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SSmartyPants posted the thread High School Taking a limit and getting the wrong answer...don't know why in Calculus.The limit in question: $$\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} = 2$$ As stated in the TL;DR, I understand...