Question

What is the limit of (x Sin (1/x)) as x becomes positive infinite?

Answer 1

i believe the limit is 0 because as x increases, 1/x decreases to where it is approaching zero and the sin (0) is 0. and anything multiplied by 0 is 0.

Answer 2

Use a substitution u=1/x, then the limit will be
\[\lim_{u \rightarrow 0}{\sin x \over x}=1\]

Answer 3

it is 0

Answer 4

Sorry
\[\lim_{u \rightarrow 0} {\sin u \over u}=1\]

Answer 5

limits is zero , its seen by pinching theorm

Answer 6

Anwar..i think you might have to use L'Hopital's Rule.

Answer 7

I don't think L'hopital's rule can work here. I still believe it's 1 by the substitution I used.

Answer 8

check it, sub in a big number'

Answer 9

it goes to zero

Answer 10

the larger you're number gets, the closer you are to zero because it is a fraction.

Answer 11

@thamir: When substituting \(u=1/x\), \(\sin (1/x)= \sin u\) and \(x=1/u\). Also the approaching point will be zero instead of infinity since \(1/\infty \rightarrow 0\).

Answer 12

-1<sin(1/x) < 1 , this is known , multiply through by x ( it is positive because we are considering x->+infinity

Answer 13

-x<xsin(1/x)<x

Answer 14

wait, thats not going to work :|

Answer 15

something similar lol

Answer 16

@elecengineer: When you substitute a large number the x goes to \(\infty \) and the sin(1/x) goes to \(0\). And \( \infty \times 0 \ne0\).

Answer 17

L'Hopitals Rule does work AnwarA and you are right the answer is 1.
lim sin(1/x)/(1/x)
=lim [(-1/x^2)cos(1/x)]/(-1/x^2)
=lim cos(1/x)
=cos(0)=1

Answer 18

\[\lim_{x \rightarrow \inf}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}\]
let u=1/x
x->inf, u->0
so the limit is 1

Answer 19

Thanks AnwarA for solving it. Thanks chris777 for teaching me the L'Hopitals Rule. And last but not least thanks to elecengineer for trying to solve using the squeeze theorem.

Now, I need to get the correct answer which is 1 by using the squeeze theorem. When trying I reached the same point where elecengineer reached. How can we solve it using the squeeze theorem?