Question

lim a->3+ of (sin(pi/2(u^3-6u-6)))/(u-1)

Answer 1

If i just plug in 3 i get -1/2......
Is that right?

Answer 2

There are no a's in your expression?!!

Answer 3

And it's also hard to read. You should rewrite it using the equation feature.

Answer 4

oh, replace the u's with as

Answer 5

\[\lim_{u \rightarrow 3} \sin(\pi/2(u^{3}-6u-6))/(u-1)\]

Answer 6

SATELLITE, i just plugged in 3, and I got -1/2... Is that all i have to do??

Answer 7

\[\lim_{u\rightarrow 3}\sin(\frac{\pi(u^3-6u-6)}{2(u-1)})\]??

Answer 8

kno, it's pi/2 all in the numerator

Answer 9

no*

Answer 10

this thing should be continuous for all values of u except 1, so whatever you get when you plug in u = 3 is the answer

Answer 11

Okay, well I plugged in 3 and got -1/2. But wolfram isn't getting that, it's distributing the pi/2. Are you sure?

Answer 12

let me calculate.
\[\frac{3^3-6\times 3-6}{3-2}=\frac{3}{2}\]

Answer 13

\[\frac{\pi}{2}\times \frac{3}{2}=\frac{3\pi}{4}\]

Answer 14

and
\[\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}\]

Answer 15

No you rewrote it wrong. The 2 is not in the denominator.

Answer 16

hold the phone
a fraction only has a numerator and a denominator. so the 2 is in the denominator

Answer 17

So... U^3-6u-6=27-24=3
sin((pi/2)*3)=sin(3pi/2)=-1

So -1/2

Answer 18

Oh wow, so that 2 IS on the bottom? that changes everything

Answer 19

\[\frac{\frac{\pi}{2}(u^3-6u-6)}{(u-1)}=\frac{\pi(u^3-6u-6)}{2(u-1)}\]

Answer 20

Yeah. I had a very bad trig teacher and I never learned most of the trig.

How did you figure out sin(3pi/4)=sqrt(2)/2

I can figure out 3pi/2, because I can draw the sin curve, but I can't figure out 3pi/4