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#
