Limit as x approaches zero of (sec(3x)-1)/3x. Suppose secX=a[0]+a[1]X+a[2]X^2+a[3]X^3+..., a power series in X, where the various a values are coefficients. When X=0, secX=1 because secX=1/cosX, and cosX=1 when X=0, therefore a[0]=1.
Now differentiate secX with respect to X. This is the same as differentiating (cosX)^-1=-(cosX)^-2*(-sinX)=secXtanX.
If we differentiate the power series we get a[1]+2a[2]X+3a[3]X^2+... =secXtanX. When X=0, tanX=0 so a[1]=0. The power series for secX becomes 1+a[2]X^2+... so far.
Differentiate again: tanX(cosX)^-1 becomes the derivative (secX)^2(cosX)^-1+tanXsecXtanX=(cosX)^-3+secX(tanX)^2. This expression is 1 when X=0, so the next derivative of the power series is 1 when X=0: 2a[2]=1 so a[2]=1/2 and the series for secX becomes 1+X^2/2+... so far. If we replace X by 3x we have 1+9x^2/2, so sec3x-1=9x^2/2 to the second degree. Divide this by 3x and we get 3x/2. So this is the function of x we need to look at to find out what happens as x approaches zero, so now we can see that 3x/2 also approaches zero. (Note that if the denominator had been 3x^2 instead of 3x, the limit would not have been zero but 3/2 or 1.5.)