Question

I just started calculus and I need help solving lim x->0 sin x/ 2x^2-x

Answer 1

if we substitute x = 0 into this function we get 0 / 0 which is indeterminate
use l'hopitals rule find derivative of top and bottom

= cos x / 4x - 1

limit = 1 / -1 = -1

Answer 2

Factor out a sin(x)/x

That's one of the first limits taught. Then use the rules of limits to solve.

Answer 3

ah - yes - thats a better way to do it

Answer 4

how would you factor out a a limit?

Answer 5

Factor the denominator to x(2x-1)

Answer 6

you won't you have done l'hopitals rule yet i don't expect

Answer 7

i got that part. i just don't know what the rest is

Answer 8

\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x } \times \lim_{x \rightarrow 0}\frac{ 1 }{ 2x-1 }\]

Answer 9

what comes after that? or is that it?

Answer 10

Do you know what the first limit is? - like I said, that's quite important and is always taught early-on in beginning calculus.

Answer 11

http://www.youtube.com/watch?v=Ve99biD1KtA

Answer 12

BTW, Cindy, you don't need to know that whole proof, but MEMORIZE
\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=0\]

You'll be happy you spent the gray cells.

Answer 13

Thank you for that!