You can use Taylor series. I assume that is what you mean.
You can obtain the series for tan(x) by using \(\displaystyle \frac{sin(x)}{cos(x)},\)
then use their respective series, which can be found in any calc book, to find the series for tan(x)
The series for tan(x), 5 terms, is:
\(\displaystyle x+\frac{x^{3}}{3}+\frac{2x^{5}}{15}+\frac{17x^{7}}{315}+\frac{62x^{9}}{2835}\)
Subtract x and we have:
\(\displaystyle \frac{x^{3}}{3}+\frac{2x^{5}}{15}+\frac{17x^{7}}{315}+\frac{62x^{9}}{2835}\)
Now divide by \(\displaystyle x^{3}:\)
\(\displaystyle \frac{1}{3}+\frac{2x^{2}}{15}+\frac{17x^{4}}{315}+\frac{62x^{6}}{2835}\)
It isn't hard to see what the limit is now as x approaches 0.
