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My professor wrote the following on the board: $$\lim_{x\to 9\\y\to 1}\frac{\tan(y-1)\sin^2(x-9y)}{(x-9)^2+(y-1)^2}=\lim_{x\to 9\\y\to 1}\frac{\tan(y-1)}{(y-1)}\frac{\sin^2(x-9y)}{(x-9y)^2}\frac{(y-1)(x-9y)^2}{(x-9)^2+(y-1)^2}$$ When I asked him how he went from what's on the left side to the right side he answered "I just multiplied and divided by the same value" does anyone know to what value he is referring to?

Note: you can ignore the lim

May someone give a clear answer?

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  • $\begingroup$ Note that the $\lim$ should be $\lim_{x \to 9, y \to 1, x \neq 9y}$ if you want to expand it to the right hand side. $\endgroup$ – copper.hat Jun 5 at 22:52
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Multiply & divide by $(y-1)(x-9y)^2$ to given expression & rearrange the terms as follows $$\lim_{x\to 9\\y\to 1}\frac{\tan(y-1)\sin^2(x-9y)}{(x-9)^2+(y-1)^2}\times \frac{(y-1)(x-9y)^2}{(y-1)(x-9y)^2}$$ $$=\lim_{x\to 9\\y\to 1}\frac{\tan(y-1)}{(y-1)}\frac{\sin^2(x-9y)}{(x-9y)^2}\frac{(y-1)(x-9y)^2}{(x-9)^2+(y-1)^2}$$

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It's just this: $(y-1)(x - 9y)^2$

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  • $\begingroup$ could you elaborate, still I don't see how this reaches to the right side $\endgroup$ – Daniel98 Jun 5 at 22:42
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    $\begingroup$ @Daniel98 Look at the numerator: it is pretty obvious that it is being multiplied by $(y-1)(x-9y)^2$. In the denominator, the $y-1$ and the $(x-9y)^2$ are just split into the first and second fractions' denominators. $\endgroup$ – Aiden Chow Jun 5 at 22:49

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