Question

Using l'hopital's rule: lim x→∞ [x^2 sin(1/x)]?

Answer 1

Do you know what l'hopital's rule is?

Answer 2

yes , i arrange the function to sin 1/x over 1/x^-2 to get 0/0

Answer 3

but then after that, i take the derivative of both of them , but end up with 0 / infinity

Answer 4

You can do L'hopital's rule multiple times until you get an answer, assuming what you're doing is right to begin with.

Answer 5

but i must get 0/0 or infinity/infinity to continue

Answer 6

Just take the derivative of the original function, and see what you get when x goes to infinity.

Answer 7

What's 0/infinity equal to? Is it 0?

Answer 8

i did , and the limit for that derivative is 0 over infinity

Answer 9

That would be 0, but that wouldn't be the correct answer.

Answer 10

yea..i saw the answer is 1

Answer 11

http://www.wolframalpha.com/input/?i=%20lim%20x%E2%86%92%E2%88%9E%20x%5E2%20sin(1%2Fx)&t=crmtb01

Answer 12

for the derivative of sin (1/x), isnt it -1/x^2 (cos(1/x))?

Answer 13

then this limit will be 0

Answer 14

Well I looked at it like this:
\[\frac{ \sin x^{-1}}{ x^{-2} }\]\[\frac{ -x^{-2}\cos x^{-1}}{ -2x^{-3} }=\frac{ x }{ 2 }cos\frac{ 1 }{ x }\] And we can see that if we take x to infinity here we have infinity divided by 2 which is infinity and cos of 0 which is 1 times infinity. Infinity is the answer.

Answer 15

oh , but isn't x^2 does not equal to x^-2?

Answer 16

it should be 1/x^-2 right

Answer 17

If it's x^2 then it's equal to 1/(x^-2) is it not?

Answer 18

ohhhh..i think i caught my mistake

Answer 19

thanks! i put 1/ 1/ x^-2

Answer 20

Yeah it's like making a double negative lol, no problem, glad I could help.