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can you help me with this excercise.

Show using the definition of limit that

$$\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$$

Definition of limit:

$\lim_{(x,y)\to(a,b)} f(x,y) =L$ if and only if for every $\epsilon >0$ exist $\delta>0$ such that if $\sqrt{(x-a)^2+(y-b)^2}<\delta$ then $|f(x,y)-L|<\epsilon$.

Hi I´ve tried this,

For taylor series.

Given $\epsilon>0$, find $\delta>0$ such that if

$$\sqrt{x^2+y^2}<\delta$$ then $$\bigg|\frac{(1-\cos xy)\sin y}{x^2+y^2}\bigg|<\epsilon$$ \begin{align*} \bigg|\frac{(1-\cos xy)\sin y}{x^2+y^2}\bigg| & = \bigg|\frac{(1-1-\frac{x^2y^2}{2}+\frac{x^4y^4}{24})(y-\frac{y^3}{6}+\frac{y^5}{120})}{x^2+y^2}\bigg|\\ & =\bigg|\frac{(-\frac{x^2y^2}{2}+\frac{x^4y^4}{24})(y-\frac{y^3}{6}+\frac{y^5}{120})}{x^2+y^2}\bigg| \end{align*}

Then what do I do?

N. F. Taussig
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MathYobani
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1 Answers1

4

We need only use the inequality $\sin x<x$ for $x>0$ along with the trigonometric identity $\sin^2 x=\frac{1-\cos 2x}{2}$.

Then, we can write

$$\begin{align} |1-\cos xy|&=|2\sin^2(xy/2)|\\\\ &\le\frac12(xy)^2\\\\ &\le \frac14(x^2+y^2)^2 \end{align}$$

along with

$$\begin{align} |\sin y| &\le |y|\\\\ &\le(x^2+y^2)^{1/2} \end{align}$$

Therefore, we have

$$\begin{align} \left|\frac{(1-\cos xy)\sin y}{x^2+y^2}-0\right|&\le\frac{\frac14(x^2+y^2)^{5/2}}{x^2+y^2} \\\\ &=\frac14(x^2+y^2)^{3/2}\\\\ &<\epsilon \end{align}$$

whenever $(x^2+y^2)^{1/2}<\delta = (4\epsilon)^{1/3}$.

Martin Sleziak
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Mark Viola
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