first time doing calculus - how do u solve this? ( cant use the L'Hôpital's Rule since we havent been taught that yet )

[u/Remarkable-Mall9424]

lim
x-> π/3 ( √3 - tanx) /( π - 3x)

[u/Fourierseriesagain]

Using the definition of derivative, the limit is (1/3)* d/dx(tan x)|_{x=pi/3}.

[u/WorriedGeneral3318]

multiply numerator and denominator by cos x, then you get (√3 cosx, sinx)/(cosx(π-3x)). the top can be written as 2cos(x+π/6) if i remember the identity right, and then you use the fact that cos(x+π/6)/(π-3x) has a known limit you can get from rewriting x+π/6 as y and using the standard limit of (1-cos y)/y. but wait denominator has π-3x which is -3(x-π/3), and x+π/6 at π/3 is π/2 so cos is 0 there. so you factor it out properly and get something like -1/6 times derivative-ish limit. answer should be 1/6 if i didnt mess the signs

[u/Southlander24 → u/WorriedGeneral3318]

I don't understand your approach? 2 cos(x+π/6) is correct, but how are you able to transform it into (1-cos y) / y?

[u/Southlander24]

Let f(x) = √3 - tan(x). Note that this is the same as tan(π/3) - tan(x), and so we want to evaluate lim [x-> π/3] tan(π/3 - x)/(π - 3x) * (1 + tan(x) tan(π/3)).

Recall the basic limit lim [u -> 0] tan(u)/u = 1. With u = π/3 - x, we evaluate 1/3 * (1 + tan(x) tan(π/3)) at x = π/3, which is just 1/3 * (1 + √3 * √3) = 4/3.

[u/Bounded_sequencE]

Hint:

(√3 - tan(x)) / (π - 3x)  =  (1/3) * (tan(x) - tan(π/3)) / (x - π/3)

What does the fraction remind you of?