Question

If
`h(x)=(sinx)/(sin2x)`, then
`lim_(x-\pi )h(x)` is equivalent to which of the following?

a.
`-(1)/(2)`
b. 0
c. 1
d. Does not exist

Answer 1

Answer: -1/2

Explanation:

`\lim_(x \to \pi) (\sin x)/(2\sin x \cos x)\\\\=\lim_(x \to \pi) (1)/(2\cos x)\\\\=(1)/(2\cos \pi)\\\\=-(1)/(2)`

Answer 2

`-(1)/(2)`

Explanation:

If we try direct substitution we get:
`lim_(x\to\pi)h(x)=(sin(\pi))/(sin(2\pi))`

The radian
`2\pi` is a full rotation from 0 radians, in other words it's equivalent to 0 radians.

Using the unit circle as a reference you'll remember when the angle is 0 radians, the y-value is zero, so:
`sin(2\pi)=0`

When the angle is:
`\pi` it's at a half rotation where the y-value is still zero on the unit circle.

So you get the indeterminate form of:
`(0)/(0)`

We can use L'Hôpital's rule which states:
`lim_(x\to c)(f(x))/(g(x))=lim_(x\to c)(f'(x))/(g'(x))` if the original limit is in indeterminate form such as the one above.

Taking the derivative of sin(x) we simply use the property that:
`(d)/(dx)[sin(x)]=cos(x)`

For the sin(2x) we can use the chain rule which essentially states:
`h(x)=f(g(x))\\h(x)=f'(g(x))*g'(x)`

It deals with composite equations, and in this case we can represent the composite function as:

`f(x)=sin(g(x))\\g(x)=2x`

Using the chain rule we get:
`cos(2x)*2`

So now we have the limit:
`lim_(x\to \pi)(cos(x))/(2cos(2x))`

at pi radians the unit circle is at a half rotation where the x/y coordinates are:
`(-1, 0)`

so cos(pi) = -1

at 2 * pi radians were basically back at 0 radians where the x/y coordinates are:
`(1, 0)`

so cos(2 * pi) = 1

Now plugging in these values we get the fraction:
`(-1)/(2*1)` which just simplifies to:
`-(1)/(2)` which turns out to be the limit!