We are evaluating the limit:
x→0lim{tan(4π+x)}x1
This results in the indeterminate form
1∞. To resolve this, we utilize the standard exponential transformation rule:
x→alimf(x)g(x)=elimx→ag(x)[f(x)−1]
Now, we perform the subtraction
f(x)−1:
1−tanx1+tanx−1=1−tanx(1+tanx)−(1−tanx)
Simplifying the numerator, we obtain:
1−tanx2tanx
We now multiply this result by the exponent
g(x)=x1 to find the exponent of our base
e:
L=x→0lim(x1⋅1−tanx2tanx)
We rearrange the expression to isolate the standard limit
xtanx:
L=2⋅x→0lim(xtanx)⋅(1−tanx1)
Since
limx→0xtanx=1 and
limx→01−tanx1=1, the calculation simplifies to:
L=2⋅1⋅1=2