Question

UGH! limits!

Answer 1

\[\lim_{x \rightarrow \pi/2} \sin2xcosx/xsin4x\]

Answer 2

The limit of the top is 0 and the limit of the bottom is also 0, so L'hopital's rule applies.
Take the derivative of the top and bottom to get:
[2cos2xcosx -sinxsin2x]/ [4xcos4x + sin4x]
Evaluating at pi/2 gives:
[2(1)(0) - 1(0)] / [4(pi/2)(1) + (0)] =
0 / 2pi = 0

Answer 3

So, in case that seemed like gibberish, this is how you use L'hopital's rule:
1) Confirm that the original function goes to 0/0 or inf/inf or -inf/-inf. You can't use the rule if it's something like a/0 etc.
2) Take the derivative of the top and the bottom
3) Evaluate at x. If this is a determinate answer, then the limit is the same as the original function.

Answer 4

can you kind of break down how you got the derivative

Answer 5

Sure, not a problem.

Answer 6

mainly its the 2cos2xcosx that confuses me

Answer 7

Both the top and bottom use product rule which says that if I have (a*b)', the way to get it is a'*b + b'*a.
For the top that gives sin(2x)*cosx
will be (sin2x)'*cosx + (cosx)'*sin(2x)

Answer 8

right that part makes complete sense

Answer 9

Then you use chain rule. How comfortable are you with chain rule?

Answer 10

f(g(x))' = f'(g(x))*g'(x),
so for sin(2x) I get:
cos(2x)*2

Answer 11

ugh not real comfortable i get it kind of but im not real sure on when to use it

Answer 12

do i need to use if for both the sine side and the cosine side?

Answer 13

Chain rule is used whenever you have an inside function and an outside function. So sin(x) is something we know the derivative for. However, if I have something like sin(x^3), well sin(x) is the outside function and x^3 is the inside function. Basically, x^3 is the argument for sin(x).

Answer 14

ok that actually clears up alot of confusion

Answer 15

The hard part about chain rule is recognizing the outside and inside functions. Once you see which is which, the derivative is just:
[derivative of outside function]*[derivative of inside function]

Answer 16

so then at the end for thsi problem you just sub in pi/2 in for x everywhere

Answer 17

Yes.

Answer 18

Basically, if I have f(x)/g(x) and x->c
the first thing to check is that f(c)/g(c) = 0/0 or some other indeterminate form.
and then I can say that the lim is f'(c)/g'(c)

Answer 19

so (-sin(2(pi/2))sin(pi/2)+2cos2(pi/2))/(4(pi/2)cos4(pi/2) +sin4(pi/2)

which is (-sin(pi)(1)+2cos(pi))/(2picos(2pi)+sin(2pi))

Answer 20

((0)(1)+2(1))/(1(2)+(0).... my math cant be right because it should be 0 right?

Answer 21

I believe the derivative you have on top is incorrect.
You have -sin2xsinx + 2cos2x
it should be -sin2xsinx + 2cos2xcosx

Answer 22

Everything else looks right. Fix that and you're good to go.

Answer 23

haha so i do