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Evaluate the limit: $\lim_{x\to 0} x.\sin \left(\dfrac {1}{x} \right)$

My Attempt:

We know, $$-1\leq \sin \left(\dfrac {1}{x}\right) \leq 1$$ Multiplying each term by $x$ $$-x\leq x \sin \left(\dfrac {1}{x}\right) \leq x$$ for $x>0$

$$-x\geq x \sin \left(\dfrac {1}{x}\right) \geq x$$ for $x<0$

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  • $\begingroup$ That's good, now what happens as $x$ gets closer to zero. $\endgroup$ – Rab Sep 21 '17 at 0:22
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    $\begingroup$ While your approach is good, you may also observe that a substitution $x=1/t$ leads you to an easier limit for which the squeeze theorem is not needed $\endgroup$ – imranfat Sep 21 '17 at 0:41
  • $\begingroup$ @imranfat: I really don't get how $x=1/t$ helps here. Can you please elaborate further? $\endgroup$ – Paramanand Singh Sep 21 '17 at 6:21
  • $\begingroup$ It is much simpler to apply the definition of limit here and show that the limit is $0$. Your approach needs to be completed via application of Squeeze Theorem. $\endgroup$ – Paramanand Singh Sep 21 '17 at 6:22
  • $\begingroup$ @ParamanandSingh You get $\frac{sint}{t}$ with $t$ going to infinities. A bounded numerator divided by an unbounded denominator $\endgroup$ – imranfat Sep 21 '17 at 16:35
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You have the right approach, this is the canonical example of the squeeze theorem.

Take care to approach zero from both negative and positive sides and conclude that their limits are the same.

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