Thursday, November 29, 2018

Limit of goniometric function without l'Hospital's rule




I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?



lim


Answer



Set t=\frac \pi2 - x,
\lim_{x\to {\pi\over 2}} \frac {1-\sin x}{(\frac\pi2 -x)^2}=\lim_{t\to {0}} \frac {1-\cos t}{t^2}=\lim_{t\to {0}} \frac {2 \sin ^2(t/2)}{4(t/2)^2}={1\over 2}


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