Question

lim as x approaches 0 of..
(sin^5(2t))/(6t^5)

Answer 1

I haven't been taught that yet though :/ I suppose we're supposed to be do it algebraically I just haven't figured it out.

Answer 2

It won't work anyway. Anyway, you do know \(\displaystyle \lim_{x\rightarrow0} \dfrac{\sin x}{x}\), right?

Answer 3

I think you need to know L'Hopital Rule. You only need to use it once, though. I don't see how it can be solve algebraically without the use of L'Hopital Rule.

Answer 4

Can you use graph calculator to solve this?

Answer 5

Sorry, went to go eat haha. but I don't see why not? Instructions didn't prohibit it.

Answer 6

as for the limit of sinx/x, =1?

Answer 7

Using the calculator the limit seems to be 0? Doubt I'll get any work credit for just stating that though.

Answer 8

Yeah. hold on. We want to make it simply, right? So notice that both numerator and denominator has same power, right?

So we can do this:
\[\lim_{x\rightarrow 0} \dfrac{\sin^5 (2x)}{6x} = \left(\lim_{x\rightarrow 0}\dfrac{sin(2x)}{\sqrt[5]{6} ~ x}\right)^5 = \left(\dfrac{1}{\sqrt[5]{6}}~\lim_{x\rightarrow 0}\dfrac{sin(2x)}{x}\right)^5\] Right?

Now graph \(\dfrac{sin(2x)}{x}\). What do it approaches to as x approaches to zero?

Answer 9

Oooo, 2!
Would this be the answer?

Answer 10

I do still have that constant outside, do I multiply the limit times the constant?

Answer 11

Yeah, we are not done yet. Now substitute this ugly and horrible limit to nice and handsome two.
\[\cdots = \left(\dfrac{1}{\sqrt[5]{6}}\cdot 2\right)^5 =~ ?\]

Answer 12

Yes.

Answer 13

Ah, gotcha. Thanks a bunch

Answer 14

You're welcome.