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How do I solve $\displaystyle \lim_{x \to 0} \sqrt{x^2 + x^3} \sin \frac{\pi}{x}$ using squeeze Theorem?

My book only teaches me the simplest use of the Theorem. I have no idea what should I do with a function as complex as this...

I know I have to start with:

$$ -1 \le \sin \frac{\pi}{x} \le 1$$

But what do I do next?

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    $\begingroup$ Multiply both sides for $\sqrt{x^2 + x^3}$ $\endgroup$
    – rlartiga
    May 11, 2014 at 1:45
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    $\begingroup$ It seems as if at least fifty-percent of questions posted here misuse the word "solve" in this way. One solves problems. One solves equations. One does not "solve" expressions; one evaluates them. $\endgroup$ May 11, 2014 at 1:49
  • $\begingroup$ @rlartiga : By "both" I surmise that you mean "all three". $\endgroup$ May 11, 2014 at 1:50
  • $\begingroup$ @MichaelHardy yes lol :-) $\endgroup$
    – rlartiga
    May 11, 2014 at 1:51

2 Answers 2

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Then you could say $$-\sqrt{x^{2}+x^{3}}\le \sqrt{x^{2}+x^{3}} \, \sin\left(\frac{\pi}{x}\right)\le \sqrt{x^{2}+x^{3}}.$$ The outermost functions approach the desired limit, and so by the Squeeze Theorem, you get the desired result.

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  • $\begingroup$ Which would be 0, right? $\endgroup$ May 11, 2014 at 1:45
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    $\begingroup$ @user3347814 Yes! $\endgroup$
    – Ivo Terek
    May 11, 2014 at 1:52
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You have $$-1 \leq \sin \frac{\pi}{x} \leq 1$$

Multiply it by what is left:

$$-\sqrt{x^2 + x^3} \leq \sqrt{x^2 + x^3} \sin \frac{\pi}{x} \leq \sqrt{x^2 + x^3}$$

What happens when $x \to 0$? The limit we want will be squeezed between which values?

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