Thread: L'Hospital's Rule Problem

My professor did this problem on the board, but now I realize that I don't understand it:

--> The limit as x goes to 0 of sin(3x)/tan(4x)

It is a 0/0 problem, so it seems like I would apply l'Hospital's Rule and get 3cos(3x)/4sec^2(4x). But instead, she wrote:

1. lim sin(3x)*(1/(sin(4x))*(cos(4x)) --This is just separating the equation, but I don't know why she did it instead of using l'Hospital's Rule directly.

Then she wrote to RECALL that the limit as x goes to 0 of sin(Ax)/Ax using l'Hospital's Rule goes to Acos(Ax)/A, which equals 1. This makes sense.

Then:

2. lim = [sin(3x) (1/sin(4x)) (cos(4x))] 3x/3x * 4/4

3. lim = sin(3x)/(3x) * 4x/sin(4x) * 3/4 cos(4x)

4. and the limit of sin(3x)/(3x) = 1, the lim of 4x/sin(4x) =1, and the lim of 3/4 cos(4x) = 3/4, so the final answer is 3/4.

I understand why sin(3x)/(3x) and so on would have a limit of 1, due to what she told us to recall. But I really don't understand at all why she did what she did and how she did steps 2 and 3. Please help me. Thanks.

It is simple.She wanted the question to be solved without using L'Hospital's Rule.

Seems like she had sin(3x)/sin(4x) (plus some other stuff) and she wanted to reduce this. So she multiplied by "1" twice, first by 3x/3x and then by 4/4 in order to get two patterns of sin(Ax)/Ax and Ax/sin(Ax).

So she had sin(3x)/3x and 4x/sin(4x) which are both one. This leaves a 3 on top and a 4 on bottom. Do you see that? So plus the "other stuff" it is all together (3/4)*cos(4x). Take the limit as x goes to zero and that cos(4x) becomes a 1. So the final answer is just 3/4.

Patrick

EDIT: pankaj is correct that this does not use L'Hospital's Rule. Maybe that is what was throwing you off.