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How would I find the limit of the function $\frac{x^2-4y^2}{x+2y}$ as $(x,y) \to (2,-1)$. So far I have considered lines approaching the point from different direction like $x=-2y$ but every time I substitute it into the function I get $0/0$. Does this mean the limit doesn't exist? Many thanks for your help

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Hint $$x^2-4y^2 = (x-2y)(x+2y)$$

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  • $\begingroup$ How did I miss that thanks a lot $\endgroup$ – Thomas Jan 30 '17 at 21:22
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The most important reason for the existence of the concept of "limit" is to deal with limits in which the numerator and denominator both approach $0$ and the limit exists. If limits of this kind didn't exist, then derivatives wouldn't exist, because derivatives are limits of expressions in which the numerator and denominator both approach $0$. $$ \begin{align} \text{If } f(x) & = x^3 \\ \text{then } f'(x) & = \lim_{h\to0} \frac{(x+h)^3 -x^3} h = 3x^2 \\[10pt] & \text{ This } \uparrow \text{ is a limit in which the numerator and denominator both approach 0.} \\ & \text{This limit does exist. It is } 3x^2. \end{align} $$

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