SmartyPants's latest activity

Yes, I went ahead and calculated the limit using Taylor series the other night...it's alot easier this way than trying to simplify...

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 =...

Here's something interesting, and possibly useful: In this case, we needed ##t'''(0)##, where ##t(x) = \tan x##. Note that: $$t'(x) =...

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) =...

Your step:$$\lim_{x\to0}\frac{\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin...

I would have been tempted to try letting ##y = \sin x##. I don't know if that helps.

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.

Ah, I see. So alternatively, the limit of the product is NOT the product of the limits when either ##f(x)## or...

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...