What is $\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}$ for $a\ne 0$? I have found that the limit coould be $\frac{3a}{2}$ to proof it I have come to the point where $\left|\frac{x^3-y^3 }{x^2-y^2}-\frac{3a}{2}\right| = \left|\frac{x^2+xy+y^2 }{x+y}-\frac{3a}{2}\right|\le \left|x+y-\frac{3a}{2}\right|$. I have to go to something like $\sqrt{(x-a)^2+(y-a)^2}$ however I have no idea what to do from here.
Whats is the limit of $\lim_{(x,y)\to(a,a)}\frac{x^3-y^3}{x^2-y^2}$ for $a\ne0$?
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$\begingroup$
calculus
limits
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1I have used that as you might see – 2017-02-15
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0Yeah, brain death on my part. – 2017-02-15
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0Once you reduce the fraction you should just apply the appropriate limit laws. – 2017-02-15
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1It's not clear how you get your inequality, nor why it matters. It is not true that if $U
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1In fact, the inequality is false. – 2017-02-15
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0According to my book i have to show that for a $\delta = \delta(\epsilon)$ the following is true $\left|f(x,y)-L\right|\le\epsilon$ and that $\sqrt{(x-a)^2+(y-b)^2} \le \delta$ for L as the value of the limit and (a,b) being the point of the lmit (in my case this is (a,a). That is the reason whhy have used that inequality – 2017-02-16
1 Answers
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$$\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{(x-y)(x^2+xy+y^2) }{(x-y)(x+y)}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^2+xy+y^2}{x+y}=\frac{3a}{2}$$
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0According to my book i have to show that for a $\delta = \delta(\epsilon)$ the following is true $\left|f(x,y)-L\right|\le\epsilon$ and that $\sqrt{(x-a)^2+(y-b)^2} \le \delta$ for L as the value of the limit and (a,b) being the point of the lmit (in my case this is (a,a) – 2017-02-16