# Prove that limit (x^2)*[sin(1/x)]=0, x->0?

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`lim_(x->0) x^2 sin (1/x)=lim_(x->0) x lim_(x->0) sin(1/x)/(1/x)=1 lim_(x->0) x=0`

To prove that lt x-->0 {x^2)[sin(1/x)] = 0.

Proof:

Method 1: |sin theta| < 1, for any theta.

So Lt x--> 0 x^2 {sin(1/x)} should belong to (-x^2 , x^2) as x-->0, which is obviously zero .

Second method:

Consider a circle of radius with P on the circumference, centre O , making an angle theta with the initial line OX. Or angle XOP = theta. Let PN be the perpendicular from P to OX. Then for any angle theta,

PN < arc length (for the angle ) along the circumference.

O r sin theta < r theta. Or

sin theta < theta. Or

sin (1/x) < 1/x . Mutiplying by x^2,

x^2 sin (1/x) < (1/x)x^2 = x.

Now as x-->0, x^2 sin(1/x) < x = 0.

For the beginning, we'll have to make the remark that we can't use the product law here (the limit of a product is the product of limits).

This thing happens because of the fact that lim sin(1/x) does not exist, when x tends to 0.

Though, -1 =< sin (1/x) =< 1

Now, if we'll multiply the inequality above, by x^2, because x^2 is positive, for any value of x, the inequality remains unchanged.

-x^2 =< (x^2)*sin (1/x) =< x^2

If we'll calculate lim x^2 = lim -x^2 = 0.

Now, we'll apply the Squeeze Theorem and we'll get :

lim -x^2=< lim (x^2)*sin (1/x) =< lim x^2

0=< lim (x^2)*sin (1/x) =<0

So, lim (x^2)*sin (1/x) = 0.

The evaluation of the limit is the one given by the enunciation.

The limit `lim_(x->0) (x^2)*sin(1/x)` has to be determined.

If `y = 1/x` , as the value of x increases, the value of y decreases and conversely as x decreases, the value of y increases.

If `y = 1/x` and x tends to 0, y tends to `oo` .

In determining `lim_(x->0) (x^2)*sin(1/x)` , we don't need to consider what the value of sin(1/x) becomes as x tends to 0, as the product of 0 and any number is 0. The square of x as x tends to 0 also tends to 0.

This gives:

`lim_(x->0) (x^2)*sin(1/x)`

= `0^2*sin(1/0)`

= 0

The limit `lim_(x->0) (x^2)*sin(1/x)` is equal to 0.