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I'm having trouble in finding multi-variable limit and hope that an example like this could get me started on my work $$\lim \limits_{(x, y) \to (2,0)} \frac{1-cosy}{xy^2}$$

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It is a limit in two variable but the indeterminate form that comes out is due to a one of them. Then we calculate: $$\lim \limits_{(x, y) \to (2,0)} \frac{1-\cos y}{xy^2}=\frac{1}{2}\lim_{y\to 0}\frac{1-\cos y}{y^2}$$ But it is a well-known limit: $$\lim_{t\to 0} \frac{1-\cos(t)}{t^2} = \frac{1}{2} $$ Then $$\lim \limits_{(x, y) \to (2,0)} \frac{1-\cos y}{xy^2}=\frac{1}{2}\lim_{y\to 0}\frac{1-\cos y}{y^2}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$$

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  • $\begingroup$ Would I do something similar like plugging in x or y and pulling it out of the limit to most other limits? $\endgroup$ – Kevin Jul 16 '15 at 5:48
  • $\begingroup$ No, there is not a general rule. It depends on the limit. $\endgroup$ – Mark Jul 17 '15 at 17:54
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$1-\cos(y)=2 \sin^2 \left(\frac{y}{2}\right)$

Then use the equivalency of $\sin^2\left(\frac{y}{2}\right)\sim \frac{y^2}{4}$ when $y$ is close to $0$.

The answer is $\frac{1}{4}$

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    $\begingroup$ It would be beneficial to use TeX code going forward. You will find a tutorial here. $\endgroup$ – Mark Viola Jul 16 '15 at 5:49

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