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I want to find this limit: $$\lim\limits_{(x,y)\to(0,0)} \frac{x^4-y^4}{x^2+y^2}$$

Approaching the point $(0,0)$ from the x-axis, I set $y=0$ then evaluate the limit of the resulting one-variable function: $$\lim\limits_{x\to0} \frac{x^4}{x^2}=0$$ Same for the y-axis: $$\lim\limits_{y\to0} -\frac{y^4}{y^2}=0$$ So, if the limit exists, it must be $0$.

Then I invoke the delta-epsilon definition of a limit.

Let $\epsilon>0$. There must exist $\delta>0$ such that if $0<\sqrt{x^2+y^2}<\delta$, then $|\frac{x^4-y^4}{x^2+y^2}|<\epsilon$.

So, since $\delta$ must be expressed in terms of $\epsilon$, we are supposed to re-express the second inequality such that it has the same form as the first inequality, then set $\delta$ to the appropriate quantity in terms of $\epsilon$.

Accordingly, $$|\frac{x^4-y^4}{x^2+y^2}|<\epsilon\\ \frac{|x^4-y^4|}{x^2+y^2}<\epsilon\\ |x^4-y^4|<\epsilon(x^2+y^2)\\ \frac{|x^4-y^4|}{\epsilon}<x^2+y^2\\ \frac{1}{\epsilon}<x^2+y^2\\ x^2+y^2>\frac{1}{\epsilon}\\ \sqrt{x^2+y^2}>\sqrt{\frac{1}{\epsilon}}$$

Obviously this re-expression does not work out because I can't express $\delta$ in terms of anything, but I don't see how else I can re-express this. Any help?

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Switching to polar coordinates, we get $$\lim_{r\to 0}{\frac{r^4\left(\cos^4(\theta)-\sin^4(\theta)\right)}{r^2}}=\lim_{r\to 0}{r^2\left(\cos^4(\theta)-\sin^4(\theta)\right)}=0$$ regardless of the value of $\theta$.

In these problems, the answer is generally zero when the degree of the numerator is larger than that of the denominator, and indeterminate otherwise - polar coordinates is usually the way to go.

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  • $\begingroup$ Could you briefly explain how to convert to polar coordinates? $\endgroup$ – user1641398 Mar 24 '14 at 3:34
  • $\begingroup$ @user1641398 You can use the formulas $x=r\cos(\theta)$ and $y=r\sin(\theta)$. $r$ represents the distance of a point from the origin, and $\theta$ is its angle from the x-axis. Note that I factored in the numerator of the limit after converting to polar. $\endgroup$ – user137500 Mar 24 '14 at 3:36
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Simple algebra (which perhaps is what Lacarguy's answer meant but I can't understand):

$$\frac{x^4-y^4}{x^2+y^2}=\frac{(x^2+y^2)(x^2-y^2)}{x^2+y^2}=x^2-y^2\xrightarrow[(x,y)\to(0,0)]{}0$$

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The limit is zero.

$x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) \Rightarrow \frac{x^4 -y^4}{x^2 + y^2}= x^2 - y^2 < x^2 + y^2$.

So let $\epsilon > 0$ be given, choose $\delta = \epsilon^2 > 0$ then the result follows.

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  • $\begingroup$ I don't see how to apply those equations to the problem. $\endgroup$ – user1641398 Mar 24 '14 at 3:38

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