Question

what is the limit as x approaches 0 of sin^2x/x

Answer 1

what method do you have at your disposal?

Answer 2

graph, then prove algebraically, I could subsitute in 0.. but that just makes it a mess

Answer 3

it's sin^(2)x btw

Answer 4

actually we do not need much
\[\frac{\sin^2(x)}{x}=\sin(x)\times \frac{\sin(x)}{x}\]

Answer 5

as \(x\to 0\) we get \(\sin(x)\to\sin(0)=0\) since sine is continuous we can just plug in the number

Answer 6

and a well known limit is that
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]
so we get \(0\times 1=0\)

Answer 7

here is a nice picture, you can see it goes right through \((0,0)\)
http://www.wolframalpha.com/input/?i=sin^2%28x%29%2Fx

Answer 8

I'm confused about as x approaches 0 we get sinx to sin0, that confuses me.

Answer 9

ok lets go slow

Answer 10

most all function you know are continuous on their domains. that is to say, if you want to compute a limit of \(f\) as say \(x\to a\) you would simply compute \(f(a)\)

Answer 11

if \(a\) is not in the domain, then you have to do something else
but for example if i want
\[\lim_{x\to 2}\frac{x^2+x}{x+1}\] i would compute
\[\frac{2^2+2}{2+1}\] and be done with it

Answer 12

now you have
\[\sin(x)\times\frac{\sin(x)}{x}\]and you want the limit as x goes to zero
0 is not in the domain of this thing, because you have \(x\) in the denominator
but one thing you probably have seen is that
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

Answer 13

and by what i wrote above, since \(\sin(x)\) is continuous everywhere, we know
\(\lim_{x\to a}\sin(x)=\sin(a)\) so in particular \(\lim_{x\to 0}\sin(x)=\sin(0)=0\)

Answer 14

in other words the limit of the function is equal to the value of the function
now armed with the knowledge that \(\lim_{x\to 0}\frac{\sin(x)}{x}=1\) and also knowing that "the limit of a product is the product of the limits" we get
\[\lim_{x\to 0}\sin(x)\times \frac{\sin(x)}{x}=0\times 1=0\]