Wednesday, 3 September 2014

calculus - limxto0fracsinxxx2 without L'Hospital or Taylor



It is easy to see that limbut I can't figure out for the life of me how to argue without using L'Hospital or Taylor. Any ideas?


Answer



In THIS ANSWER, I used the integral definition of the arcsine function to show that for 0 \le x\le \pi/2, we have the inequalities




x\cos(x)\le \sin(x)\le x \tag 1



Using the trigonometric identity 1-\cos(x)=2\sin^2(x/2), we see from (1) that



-2x\,\,\underbrace{\left(\frac{\sin^2(x/2)}{x^2}\right)}_{\to \frac14}\le \frac{\sin(x)-x}{x^2}\le 0 \tag2



Applying the squeeze theorem to (2) yields the coveted limit





\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\sin(x)-x}{x^2}=0}



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