Lets first try plugging in the value for the limit since from general limit rules if the function is continuous we can do such.
limx->0 (2tan(0))/(0(sec(0)) well we know that the bottom number will give us the value of 0 , but what about the top?
well, since tanx = sin(x)/cos(x) and sin(0) = 0 and cos(0) = 1 then we are left with a solution of limx->0 (2tan(0))/(0(sec(x)) )= 0/0.
We know from the L'hospital rule. That when we have a limit of 0/0 we can find the derivative of the top function and bottom function. Then plug in the limit.
Therefore using l'hospital rule. Finding d/dx of the numerator and denominator
limx->0 (2tan(x))/(x(sec(x))
d/dx 2(tanx) = 2(sec^2(x))
d/dx (x(sec(x)) = sec(x) + xtanxsecx
now, limx->0 (2(sec^2(x)))/(sec(x) + xtanxsecx)
limx->0 (2(sec^2(0)))/(sec(0) + (0)tan(0)sec(0))
sec(0) = (1)/(cos(0)) = 1 recall the unit circle where cos(0) = 1
Now we know that the zero in (sec(0) + (0)tan(0)sec(0)) will cancel out the second term of (0)tan(0)sec(0)) , so the denominator equals 1
Let us find the numerator, well we know that sec(0) = (1)/(cos(0)) = 1 and recall sec^2(x) = (secx)^2, so
(2(sec^2(0))) = 2(1) or simply 2
Now, this means that limx->0 (2(sec^2(0)))/(sec(0) + (0)tan(0)sec(0)) = 2/1 or 2