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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

User Adrisons
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2 Answers

27 votes
27 votes

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)

User Utensil
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2.8k points
24 votes
24 votes

Answer:


-(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!

User Arup Bhattacharya
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2.8k points