00:14
Okay, we are told in the directions for this problem that this is the indeterminate form of infinity over infinity.
00:20
So we will use lebetal's rule.
00:27
So the derivative of the numerator, tangent, is sikin squared.
00:35
And the denominator derivative, we can use the quotient rule.
00:42
So the quotient rule is low, the denominator.
00:46
Oops, that's 2x.
00:51
D high derivative of the numerator 0 minus high, d low derivative of the denominator is 2 all over the denominator squared.
01:05
All right, so i'd like to simplify this, and we have sequence squared numerator divided by, and then i'm going to write the denominator.
01:13
So that first part is nothing, so that's negative 6 over 2x minus pi squared.
01:21
Now, i'm going to rewrite sikin.
01:23
Seekin is the reciprocal of cosine.
01:26
So i'm going to make this 1 over cosine x squared, which i'm going to write like that.
01:32
Okay, then we can divide two fractions by flipping the second one over and multiplying.
01:47
So that's going to give us 2x minus pi squared over negative 6 times cosine x squared.
01:57
And that is the limit as x approaches pi over 2 from the left.
02:05
So let's plug this in.
02:08
So we have 2 times pi over 2 minus pi squared.
02:17
So 2 times pi over 2 with pi, pi minus pi squared is 0.
02:22
Over the denominator, negative 6 times the cosine of pi over 2 squared.
02:28
So the cosine of pi over 2 is 0.
02:32
So we get 0 over 0, and we have to use the local 12 again.
02:38
So we're doing the derivative right here...