Find the limit --> 0

Question

Find .

Question options:

1)	0
2)	Does not exist
3)
4)

Answer 1

Explanation:

The key fact that you need to know to do this problem is: lim x→0 sin(x)/x = 1.

It's also true that lim x→0 sin(5x)/(5x) = 1; we could replace 5 with any non-zero number and this would still be true.

Lastly, lim x→0 x/sin(x) = 1. (We will use this later)

When evaluating the limit of a quotient, it's a good idea to start by evaluating the limit of the numerator and denominator. In this case we get lim x→0 sin(-3x) = 0 and lim x→0 sin(2x) = 0 which gives us 0/0 which is what's called an indeterminate form. This means we need to be more careful.

Full Answer:

Here's the algebra trick that gets you there:

lim x→0 sin(-3x)/sin(2x)

[step 1] = lim x→0 ( ( sin(-3x)*(-3x) ) / (-3x) ) / ( ( sin(2x)*(2x) ) / (2x) )

[step 2] = lim x→0 ( sin(-3x)/(-3x) ) / ( sin(2x)/(3x) ) * (-3x)/(2x)

[step 3] = lim x→0 ( sin(-3x)/(-3x) ) / ( sin(2x)/(3x) ) * lim x→0 (-3x)/(2x)

[step 4] = 1/1 * -3/2 = -3/2

Details:

In step 1 we use the fact that sin(-3x) = sin(-3x) * (-3x) / (-3x), we do a similar thing to the denominator

In step 2 we rearrange our fraction so it's a product of two things, one of them being (-3x) / (2x). This is important because we know lim x→0 (-3x) / (2x) = -3/2.

In step 3 we use the fact that lim x→0 f(x)g(x) = (lim x→0 f(x)) * (lim x→0 g(x))

In step 4 we use the fact that sin(x)/x → 0 as x → 0 twice. (Once for the numerator, once for the denominator)

Alternate Full Answer:

If you are allowed to use L'Hospitals Rule then the following is also valid:

lim x→0 sin(-3x)/sin(2x) = (via LH Rule) lim x→0 cos(-3x)*(-3) / cos(2x)*(2) = ( lim x→0 -3*cos(-3x) ) / ( lim x→0 2*cos(2x) ) = ( -3 * cos(0) ) / (2 * cos(0) ) = ( -3 * 1) / ( 2 * 1 ) = -3/2

Answer 2

The sin of 0 is zero, so the expression resolves to 0/0, which is ambiguous.

Let's try L'Hopital's rule:

Take the first derivative of top and bottom to get:

cos(-3x)*(-3)

cos(2x)*2

cos(0) = 1, so we have -3/2

CHECK by plotting the original function