Question

lim tan(9/4pi-pisint/t)
t->0

I will post equation in comments to make it easier to see

Answer 1

\[limt \rightarrow \tan((9/4)\pi-(\pi \sin(t)/t)\]

Answer 2

I can't understand your brackets.
Lemme see if this is what you meant...

Answer 3

\[\Large\bf\sf \lim_{t\to0}\tan\left(\frac{9\pi}{4}-\frac{\pi \sin t}{t}\right)\]

Answer 4

ya

Answer 5

Let's pass the limit into the tangent ( I think we're allowed to do that ) <.<

\[\Large\bf\sf \tan\left(\lim_{t\to0} \frac{9\pi}{4}-\frac{\pi \sin t}{t}\right)\]And let's just focus on what's going on inside of the brackets for a moment,\[\Large\bf\sf \lim_{t\to0} \frac{9\pi}{4}-\frac{\pi \sin t}{t}\]

Answer 6

Recall that for small t values, \(\Large\bf\sf \sin t\approx t\)

Or you can just remember this identity if it makes more sense to you,\[\Large\bf\sf \lim_{x\to0}\frac{\sin x}{x}\quad=\quad 1\]

Answer 7

ya i remember that

Answer 8

So what do we get for this? :o\[\Large\bf\sf \lim_{t\to0} \frac{9\pi}{4}-\pi\color{royalblue}{\frac{ \sin t}{t}}\]

Answer 9

1

Answer 10

Not the whole thing, right? Just the sint/t\[\Large\bf\sf \lim_{t\to0} \frac{9\pi}{4}-\pi\color{royalblue}{\frac{ \sin t}{t}}\quad=\quad \frac{9\pi}{4}-\pi\color{royalblue}{(1)}\]

Answer 11

ya how would you simplify

Answer 12

\[\Large\bf\sf \tan\left(\frac{9\pi}{4}-\pi\right)\]Just need to get a common denominator.
Or if you remember that tangent is periodic in pi, you can probably take advantage of that to skip a step.

Answer 13

5pi/4?

Answer 14

ya sounds good.
Remember your coordinates on the unit circle? :)

Answer 15

would the answer just be tan(5pi/4) or do i need more

Answer 16

cause i don't know what tan(5pi/4) is

Answer 17

You should know! :U

Answer 18

Something to remember:
Tangent is always 1 or -1 at the pi/4 values.

Is tangent positive or negative in the 3rd quadrant?

Answer 19

positive

Answer 20

Ok good, so we simply get a positive 1.\[\Large\bf\sf \tan(5\pi/4)=1\]

Answer 21

ok theres one more really complicated one i have no clue how to solve

Answer 22

\[4\pi \sec(3e^x+2\pi \tan(3\pi/secx)-3\]

Answer 23

as x goes to 0

Answer 24

You're missing a bracket, is it kind of like this?
\[\Large\bf\sf \lim_{x\to0}4\pi \sec\left(3e^x+2\pi \tan\left(\frac{3\pi}{\sec x}\right)\right)-3\]

Answer 25

ya

Answer 26

wait the -3 at the end is inside the last bracket

Answer 27

\[\Large\bf\sf \lim_{x\to0}4\pi \sec\left(3e^x+2\pi \tan\left(\frac{3\pi}{\sec x}\right)-3\right)\]

Answer 28

ya thats it

Answer 29

This one is kind of nice.
We can simply plug in x=0 without any problem.

Answer 30

\[\Large\bf\sf \sec 0\quad=\quad \frac{1}{\cos 0}\quad=\quad ?\]

Answer 31

1

Answer 32

\[\Large\bf\sf \lim_{x\to0}4\pi \sec\left(3e^x+2\pi \tan\left(\frac{3\pi}{1}\right)-3\right)\]Ok good.

Answer 33

Any other simplification that we can make? :o

Answer 34

tan(3pi)=0?

Answer 35

wait never mind

Answer 36

Yah that sounds good.
What can we do with the e^x part?

Answer 37

\[\Large\bf\sf \lim_{x\to0}4\pi \sec\left(3e^x+2\pi(0)-3\right)\]

Answer 38

would you plug 0 in the x

Answer 39

ya

Answer 40

so thats just 3

Answer 41

so the inside is 0

Answer 42

so the answer is 4pi?

Answer 43

Mmmm ya I think that's right. I'll check it on wolfram a sec.

Answer 44

yay 4pi \c:/ gj

Answer 45

thanks!!!!!!