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

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magnet1qu3 | Student, College Freshman | eNoter

Posted May 29, 2010 at 2:33 AM via web

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

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giorgiana1976 | College Teacher | Valedictorian

Posted May 29, 2010 at 2:40 AM (Answer #1)

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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.

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neela | High School Teacher | Valedictorian

Posted May 29, 2010 at 3:00 AM (Answer #2)

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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.

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oldnick | Valedictorian

Posted May 1, 2013 at 4:56 AM (Answer #3)

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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`

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