How do I calculate the limit $$\lim_{z\to0}\frac{1-\cos z}{z^2}$$
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Taylor expand $\cos(z)$. – Matthew Cassell Jan 14 '17 at 10:09
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Related: http://math.stackexchange.com/questions/2080800/how-to-find-the-given-limits – Watson Jan 14 '17 at 10:10
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As $z\rightarrow 0$, $\cos z = 1 - \frac{z^2}{2} + O(z^3)$.
Plugging this in, we have
$$\lim\limits_{z\rightarrow 0}\frac{1-\cos z}{z^2} = \lim\limits_{z\rightarrow 0}\frac{\frac{z^2}{2}-O(z^3)}{z^2} = \lim\limits_{z\rightarrow 0}\frac{1}{2}-O(z)=\frac{1}{2}$$
Tom
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Yes , Taylor series expansion of $cos(z) = 1 - \frac{z^{2}}{2} + \frac{z^{4}}{24} - ....$,so the limit is $\frac{1}{2}$ as when divided by $z^{2}$, the terms of the Taylor series tends to 0 except the $\frac{z^{2}}{2}$ term and when divided by $z^{2}$ leaves a $\frac{1}{2}$.
Hope this helps.
BAYMAX
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