# Apply L'Hôpital's Rule on the given limit?

Hi everyone! I know why this isn't zero (because obviously 0 can't be in the denominator), but I thought maybe it is supposed to be 3, which I haven't tried yet, but when I put it into the calculator, I don't think that's right now either... can someone help me figure it out and give an explanation on why it is whatever the answer is? Relevance
• When x is small sin x = x- x^3/6+ .... Put it, You get

3x^3/(-x^3/6)=-18. You can use l'Hospital rule

3x^3/(sin x-x) take first derivative, You get 9x^2/(cos x -1). When x=0, denominator goes to 0. Take one more derivative.

-18 x/sin x. Take one more derivative. You get -18/cos x. Now put x=0. The answer is -18. This is what you get. You have to repeat till the denominator is not 0.

• L'Hopital's Rule says that if you have a limit of a fraction that is inf/inf, -inf/-inf, or 0/0, then the limit as x approaches a number is the same as the limit of the numerator's derivative over the denominator's derivative.

lim x→0 of 3x³ / [sin(x) - x]

so if we get the derivative of both halves we get:

f(x) = 3x³ and g(x) = sin(x) - x

f'(x) = 9x² and g'(x) = cos(x) - 1

So the limit of:

lim x→0 of 9x² / [cos(x) - 1]

Since it gave you the numerator, I think that it just needed you to get the expression in the denominator, even though that will also result in a denominator of 0.

Since this is still results in 0/0, we can get the derivatives again:

f(x) = 9x² and g(x) = cos(x) - 1

f'(x) = 18x and g'(x) = -sin(x)

Now we'll get:

lim x→0 of 18x / [-sin(x)]

lim x→0 of -18x / sin(x)

Which is still 0/0.  Get the derivative one more time:

f(x) = -18x and g(x) = sin(x)

f'(x) = -18 and g'(x) = cos(x)

Now we get:

lim x→0 of -18 / cos(x)

And this has a value:

-18 / cos(0)

-18 / 1

-18

But again, I think the answer the question is looking for is that first derivative:

cos(x) - 1

• so do it one more time  3 x³ / [ sin x - x ] --> 9 x² / [ cos x - 1 ] ---> 18 x / [ - sin x ] ---> - 18 since sin x / x ---> 1