Limits - Trigonometric Ex 6

Uploaded by TheIntegralCALC on 20.02.2012

Limits - Trigonometric Example 6
Today we’re going to be doing another trigonometric limits problem. And in this one, we’ve been
asked to find the limit as x approaches 0 of 1 – cos x divided by sin x, and the formula
we’re going to be using to find this limit, evaluate the limit, is cos2x + sin2x = 1.
So, with trigonometric limit problems, what we’re trying to do is manipulate the function
so that we can easily simplify and evaluate the limit. Notice that if we just plug in
the number that we’re approaching, in our case that’s 0, we get 1 – cos(0) divided
by sin(0) and if we evaluate, this cos(0) is 1 so we’d get 1 – 1 which would just
be 0 in our numerator and sin(0) is 0 so we’d end up with 0 in our denominator. This is
an indeterminate form, totally not a legitimate answer; we can’t just evaluate this limit
by plugging in the number that we’re approaching. So at this point, we have a couple of options.
We can use L’Hopital’s Rule which is the rule we use when we run into an indeterminate
form like 0 over 0, or infinity over infinity, something like that, but a lot of times when
you get to a trigonometric limits section in a course you either don’t know L’Hopital’s
Rule yet or your professor specifically wants you to evaluate using trigonometric limit
laws instead of using L’Hopital’s Rule. So, if either those things are the case we
want to know how to evaluate using trigonometric limit laws and most of the time what we try
to do is manipulate our function so that it transforms into, you know, different pieces
that are expressed in trigonometric limit laws. So, basically we’re just like converting
the function so that we can pull up particular pieces that we already know the answers to
and we can make simply substitutions from more complicated pieces. Like for example,
the fact that cos2x equal, or sorry, cos2x + sin2x is equal to 1, cos2x + sin2x looks
very complicated but we can actually just make the substitution 1 for that whole term,
so it’s kind of like that, we have this formulas, we’re trying to change our original
function into some different, into a different form so that we can make easy substitutions.
In this particular case, what we’re going to do is use the conjugate method from limits
and if you’re, you know, kind of go in order in a typical calculus course you probably
already covered conjugate method, you can check out that section on my website if you’re
not familiar with it, but basically, the conjugate of a polynomial is the same terms in the polynomial
but with the opposite sign in the middle. So, in other words, the conjugate of 1 – cos
x, our numerator, is 1 + cos x, the same two terms just flipping the sign in the middle,
literally that’s how it is, instead of 1 – cos x, 1 + cos x, that is the conjugate
of 1 – coz x. So what we’re going to do to manipulate
our function is multiply both the numerator and the denominator by the conjugate of the
numerator, there is no conjugate of the denominator because it’s only one term and we need two
terms in order to flip that sign in between two terms. So this is, multiplying both the
numerator and the denominator by the conjugate, is a fairly common technique in limits especially
in trigonometric limits to manipulate a function, so whenever you see two terms like this you
can try the conjugate method. So, the reason we do it is because it’s going to allow
us to cancel some things out in the numerator and our goal would be to cancel enough so
that when we plug in the number we’re approaching, again, we get a reasonable answer not an indeterminate
form like we got before right here. So, we’ll multiply this out and when we do we’ll get
1 + cos x – cos x – cos2x, and that whole thing is going to be divided by sin x (1 +
cos x). Just as a tip, whenever you’re doing conjugate
method, we took the conjugate of our numerator so I would multiply out completely the numerator,
because the reason we we’re doing conjugate method is in the hopes that we will be able
to cancel something, so I would multiply out the numerators completely because we’re
multiplying by the conjugate of the numerator, the denominators, I would leave separate like
I did here, right, I didn’t multiply everything together because we’re hoping to be able
to cancel something, so, don’t multiply your denominators together in this case.
As you can see in our numerator we’ve got a +cos x and and a –cos x, so these two
are going to cancel which is exactly what will happen every time when you apply conjugate
method. So what we’re left with is 1 – cos2x divided by sin x (1 + cos x). And, now we’re
going to go ahead and make substitution using our formula right here. We’ve got 1 – cos
x, so if we manipulate this formula which is just the formula from trigonometry, we’ll
subtract cos2x from both sides and we can see that what we’d get is sin2x = 1 – cos2x,
so, what that tells us is we can plug in sin2x for our numerator, and that formula from trigonometry
is very common in evaluating trigonometric limits so it’s a good one to know. So we
would plug in sin2x in our numerator and leave our denominator the same, so sin x (1 + cos
x). And as you can see now we’ve got sin2x in our numerator and sin x in our denominator
so we can cancel out one sin x term for both the numerator and denominator, and when we
do that we’re just left with one sin x term in the numerator and none in the denominator,
this goes away completely and this squared exponent goes away because we’re just left
with that one in the numerator, and then of course all we have left in our denominator
is 1 + cos x. So now, you can see that we’ve radically
changed the way that our function looks from where we started at 1 – cos x divide by
sin x, at this point we should go ahead and plug in the number that we’re approaching
again to see if we get a good answer when we evaluate at 0. So, if we plug in 0 we’ll
get sin 0 divided by 1 + cos 0. Sin 0 is just 0, and you can either evaluate that using
your calculator or the unit circle, but sin 0 is 0, and cos 0 is 1 so 1 + 1 is 2, we get
0 over 2 which is just 0. So, that’s it, that’s our final answer,
that’s a good answer. It might be deceiving because we said that this indeterminate form
here was not a good answer, 0/0, if you get 0 as your answer that’s totally fine, what
we can’t have is an indeterminate form which is the case of 0/0. But, 0 in this case is
the answer to our limit so if we write our answer, technically, we would say the limit
as x approaches 0 of the function 1 – cos x divided by sin x is equal to 0.
So that’s it, that’s our final answer. I hope that video helped you guys and I will
see you in the next one.