Question

The as x approaches 0 of tan^2x/x

Answer 1

L=Lim tan(x)^2/x x->0

Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.

d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)

d(x)/dx = 1

=>

L=2tan(x)sec^2(x)/1 x->0

= (2(0)/1^2)/1

=0/1

=0

Another way using series,

We know that tan(x) = x+x^3/3+2x^5/15+.....

then tan^2(x), using binomial expansion gives

x^2+2*x^4/3+.... (we only need two terms)

and again apply l'Hôpital's rule, we have

L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1

=0 as x->0