find the limit as x --0 (sin(kx))/(kx^2-2kx)

Question

find the limit as x -->0 (sin(kx))/(kx^2-2kx)

jamesj · Answer

Do you know the limit as x --> 0 of  sin(x) / x ?

anonymous · Answer

is it 0?

anonymous · Answer

noooo its 1!

jamesj · Answer

No, it's 1.

Using this result, you can arrange your function in such a way as you can see the result.

anonymous · Answer

how do i rearrange the function?

jamesj · Answer

$$\frac{ \sin(kx) }{kx^2-2kx} = \frac{ \sin(kx)}{kx} . \frac{1}{x - 2}$$

anonymous · Answer

take the derivative of the nominator and denominator:

kcos(kx)/(2kx -2k)

plug x = 0

k / -2k = -1/2

anonymous · Answer

i thought the answer was 1?

jamesj · Answer

Now
$$\lim_{x \rightarrow 0} \frac{\sin kx}{kx} = 1$$
Hence your limit is ...

anonymous · Answer

that one guy said the limit was (-1/2) though?

jamesj · Answer

He was talking about the entire problem.

jamesj · Answer

and he's right, but he using a rule for limits that you may not know yet.

anonymous · Answer

no no i remember that rule now but thank you

jamesj · Answer

For the record my argument is elementary:
$$\lim_{x \rightarrow 0} \frac{\sin kx}{kx^2 - 2kx} = \lim_{x \rightarrow 0} \frac{\sin kx}{kx}.\lim_{x \rightarrow 0} \frac{1}{x-2} $$
which is equal to 

1 times -1/2

I.e., -1/2.

anonymous · Answer

ok cool deal