Question

Given lim x -> c f(x)=-8, lim x->c g(x)=5, lim x->c h(x)=0.
Find lim x->c [5h(x)-2g(x)-3f(x)]

Answer 1

it would just be 5*lim(h)-2lim(g)-3lim(f) = 5*0-2*5-3(-8) = -10+24 = -14

sorry for the bad notation(lim(f)), but I thought it would help make my point clear

Answer 2

So its really just simple as that?

Answer 3

yes

Answer 4

so what's lim x->c f(x)/g(x)?

Answer 5

note that if one of those said instead of lim x->c g(x)=5 it said lim x->a g(x)=5
we could not have answered the question

Answer 6

simply
lim x->c f(x) / lim x->c g(x)

Answer 7

Alright how about this one... lim x->0 sin^2(x)/2x^2

Answer 8

the quotient of the limit is the limit of the quotient
the product of the limit is the limit of the product
the sum of the limits is the limit of the sum

Answer 9

how do I evaluate the limit if it exists?

Answer 10

you just plug it in, but here you get 0/0

Answer 11

what do you do when there is a sin^2?

Answer 12

do you know about lhopitals rule yet?

Answer 13

I've never heard of that

Answer 14

hmm, im not sure how they want you to show this...

Answer 15

Is there like a way to break down sin?

Answer 16

Or do I use the quotient rule?

Answer 17

Anybody?????????

Answer 18

no idea without series, or lhopitals...

Answer 19

some identity im sure

Answer 20

I hate trig functions

Answer 21

i just don't get it