How should I solve this exercise:
For which values of real parameter $a$ the following equality is true:
$$\lim_{x\to 0}{1-\cos{ax}\over x^2}=\lim_{x\to \pi}{\sin{x}\over \pi-x}$$
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4$\sin(x)=\sin(\pi-x)$ could help – Claude Leibovici Apr 17 '14 at 14:10
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I've tried but I haven't seen anything – wonderingdev Apr 17 '14 at 14:11
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1Try changing variables and bound in the rhs. Think also about Taylor for the lhs. – Claude Leibovici Apr 17 '14 at 14:12
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1That gives you that the RHS equals $1$. Evaluate the LHS by using L'Hôpital twice. Equating both gives a simple condition for $a$. – J.R. Apr 17 '14 at 14:12
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As pointed out by Claude Leibovici. Rewrite $$ \lim_{x\to \pi}{\sin{x}\over \pi-x}=\lim_{\pi-x\to 0}{\sin{(\pi-x)}\over \pi-x}=1.\tag1 $$ As pointed out by Your Ad Here, you will get $$ \begin{align} \lim_{x\to 0}{1-\cos{ax}\over x^2}&\stackrel{\text{l'Hospital}}=\lim_{x\to 0}{a\sin{ax}\over 2x}\\ &\stackrel{\text{l'Hospital}}=\lim_{x\to 0}{a^2\cos{ax}\over 2}\\ &=\frac{a^2}2.\tag2 \end{align} $$ $(1)$ and $(2)$ yield $\ \Large a=\large\pm\sqrt{2}$.
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