find limit(n goes to enfinity) xsin(1/x)?

Answer 1

Suppose y = 1/x,
then as 'x' goes to infinity, 'y' goes to zero:

Hence, our problem can be rewritten as:

limit (y->0) (1/y)*sin(y)

Applying L'Hospital's rule:

limit (y->0) cos(y)/1

= cos(0) = 1

Hence, the limit's value is 1.

Answer 2

Let 1/x = y

So y -->0 when x-->inf

lim xsin(1/x) as x-->inf = lim (sin y)/y as y -->0

lim (sin y)/y = 1.

Why? If y is measured in radian, the value of the sine of an angle is close to the value of the angle in radians.

let s = arc of a circle
y = angle subtended by the arc.
r = radius of the circle

The relation between them is: s = r y

For very small angle, the arc is nearly straight, so:

sin y = s/r

From above: y = s/r also.

So for very small angle (as y approaches zero):

sin y = y.

Answer 3

Any function of sin (1/x) is undefined as x tends to infinity; it oscillates faster and faster between -1 and +1.

Answer 4

lim (sin(1/x))/(1/x) = 1
n-inf