Question

@jamesJ or whoever knws, limit x->0 (2sinx-1)?? dont knw where to begin on this one..

Answer 1

use the same method as James showed you in your last question.

Answer 2

but what is sin? sin=1/cscx??

Answer 3

f(x)=2 sin(x)-1 is continuous at x=0 therefore you just evaluate
f(0) to find the limit here

Answer 4

how can you tell it is continuous?

Answer 5

because y=sin(x) is a continuous function
It exists for any real input

Answer 6

ok so on that note, is y=cosx also a continuous function?

Answer 7

yep

Answer 8

but y=tan(x) and y=sec(x) and y=cot(x) and y=csc(x) have discontinuities because at some point there bottom is zero

Answer 9

like i'm saying there functions i just mentioned can be written in terms of cos(x) and sin(x)
and remember a function is discontinuous( or one reason it can be discontinuous ) is if the bottom is zero

Answer 10

these functions* as in y=tan(x),sec(x),cot(x),csc(x)

Answer 11

u r a mind reader haha, that was exactly the question i was going to ask. thank u for ur help!!!!!