IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

joyfrancis

Find :

_[x]

_[0] x² sin(1/x)

sboosy

The answer is 0

sin (1/x) as x tends to 0 ..after all just oscillates between -1 and 1 ..which is not indeterminate form at all

so answer is 0 * some number between -1 to 1

so the answer is 0

spideyunlimited

sin x will alway be between -1 and 1, but x^2 will tend to0 so the limit is zero.

anandghegde

another way to look at it is using the sandwich/squeeze theorem
We know that $\mid \sin \frac{1}{x} \mid \le 1$

Hence $\mid x^2\sin \frac{1}{x} \mid \le x^2$

or $-x^2\le x^2\sin \frac{1}{x} \le x^2$
Taking limit as x tends to zero,

$\displaystyle\lim_{x\to\ 0}-x^2\le \displaystyle\lim_{x\to\ 0}x^2\sin \frac{1}{x} \le \displaystyle\lim_{x\to\ 0}x^2$

$0\le \displaystyle\lim_{x\to\ 0}x^2\sin \frac{1}{x} \le 0$

Hence $\displaystyle\lim_{x\to\ 0}x^2\sin \frac{1}{x} = 0$

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