Question

Evaluate the following limit if it exists. lim x->0 sin^2(x)/2x^2

Answer 1

write sin^2x / x^2 as (sin x/x)^2

and you know what lim x->0 sin x/x = ... ?

Answer 2

I don't think that's the same thing, is it?

Answer 3

lim x-> 0 (sin x/x)^2 = ( lim x-> 0 sin x/x)^2

Answer 4

yes that's equal to that... but where did you get x/x?

Answer 5

its not x/x :P its (sin x) / x ..... missed the brackets
so keeping the constant 1/2 out

1/2 lim ([sin x]/x)^2 = 1/2 ( lim (sin x)/ x )^2

Answer 6

got this ?

Answer 7

that will just =
1/2 (1)^2

Answer 8

because sin x/ x =1 when x ->0

Answer 9

so how did you get the 1/2?

Answer 10

the 2 was in the question....in the denominator
thats why 1/2

Answer 11

does [sin (x/x)] equal 1 because its set that way? Because, obviously if i plug in 0, that wouldn't give me anything, right?

Answer 12

are you gone?

Answer 13

the answer is 1/2

Answer 14

lim 1/2(sin x/x)^2
x->0

Answer 15

as sinx/x is equal to 1/therefore,the solution is 1/2.

Answer 16

sorry, i was away,
yes, lim x->0 sin x/ x is a standard limit

you can prove it in many ways , but for this Question, you just need to mention it, not prove it