Determin the limit using L'Hopital's Rule

Question

lim   1-cos(3x)/x^2

x→0

 

I took the derivative and got 3sin(3x)/2x, when I plug in zero I get 0/0

I took the derivative a second time and got 9cos(3x)/2, plugged in 0 and got 0/2

I took the derivative for a third time and got -27sin(3x)/0, which also will not work.

Answer 1

You can also solve this another way, using a trig limit. If you take the first derivatives, you get 3sin3x/2x.

 

Restate the denominator as 2/3(3x), and you now have 3/(2/3) . sin3x/3x

 

This rearranges to 9/2 sin3x/3x The limit (x →0) of sin(ABC)/(ABC) =1, so

 

THE ANSWER IS 9/2

 

Answer 2

You have the wrong answer for the second derivative since the cosine of 0 degrees is 1.