Limits - Trigonometric Ex 5


Uploaded by TheIntegralCALC on 20.02.2012

Transcript:
Limits - Trigonometric Example 5
Today we’re going to be doing another trigonometric limits problem. And in this one we’ve been
asked to evaluate the limit as x approaches 0 of the function 1/x times sin (x/3). We’re
going to be using the formula the limit as x approaches 0 of sin x divided x and that
formula tells us that if we can get our function into the form sin x divided by x we can make
the substitution of 1 for that value, that’s equal to 1.
So, with trigonometric limit problems, what you’re usually trying to do is simplify
your function or manipulate your function in to the form of one of our fundamental trigonometric
limit formulas. This happens to be the one we’re going to be using in this example.
It’s extremely common, again, there’s only two or three so this is the one we’re
using here. And, what you need to pay attention to is the fact that our formula tells us that
whatever is inside of our sine function should be equal to whatever is in the denominator.
When that is true, then we can make the substitution 1 for sin x divided by x. So what we’re
going to be primarily interested in doing is converting our function so that this is
true, but whatever is inside of our function, our sine function, will be equal to what’s
inside of our denominator. So, first, you can see that in our original
function we’ve got x/3 inside of our sine function which means we’re going to want
to get x/3 in our denominator. So first, we can manipulate our function to be the following:
sin (x/3), all divided by x, right, because this whole sine function here is multiplied
by 1/x, which means sin (x/3) can go in the numerator of that 1/x fraction, and we’ll
be left with sin (x/3) in the numerator and just this x right here in the denominator.
So that’s the first way we’re going to manipulate our function and it’s looking
closer to our formula right here, right. All we need to do now is transform this x
into x/3 and then whatever is inside the sine function and whatever is in the denominator
will match one another and we can make our substitution. So, the way that we’re going
to do that is by multiplying both the numerator and the denominator by 1/3, right, by 1/3,
because that is the way to get x/3 in the denominator. You can see that if we multiply
x times 1/3 we’d get x/3 which would match what’s inside of our sine function right
here and we’ll be able to make that substitution. So when we multiply out we’ll get 1/3 times
sin (x/3), all divided by x/3, right, 1/3 times x is x/3. So, now as you can see, this
value inside the sine function and what’s in the denominator right here match each other
so we can use our formula. So we’ll get 1/3 times, actually, let’s do this first
to simplify it, we’ll pull out this 1/3, we’ll pull that outside, and we’ll just
rewrite this as sin (x/3) divided by x/3 and now it becomes perfectly clear that this part
of our function we can use to make the substitution and plug 1 in for that. So, we’ll have 1/3
times 1 which is obviously just 1/3. And that’s it, you’re done. That’s our final answer.
So we’ll write it out technically as limit as x approaches 0 of 1/x sin (x/3) is equal
to 1/3. So again, just to recap, we transformed or
manipulated our original function so that we could pull out sin x over x, in our case
sin (x/3) divided by x/3, we pulled that out and made our substitution, plug 1 in for that
according to or formula, and that allows us to simplify to 1/3.
So, that’s it.
I hope that video helped you guys and I will see you in the next one.