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Use squeeze Theorem to evaluate the limit : $ \lim_{(x,y) \rightarrow (3,4)} (x^2-9)\cos ( \frac{1}{(x-3)^2+(y-4)^2}) $

Answer:

Let $ \ f(x,y)=(x^2-9)\cos ( \frac{1}{(x-3)^2+(y-4)^2}) $ . We know that

$ -1 \leq \cos ( \frac{1}{(x-3)^2+(y-4)^2}) \leq 1 \ \ \forall \ x,y $

or,$ -(x^2-9) \leq (x^2-9) \leq \cos ( \frac{1}{(x-3)^2+(y-4)^2}) \leq x^2-9 $

or, $ \lim_{(x,y) \rightarrow (3,4)} -(x^2-9) \leq \lim_{(x,y) \rightarrow (3,4)} (x^2-9) \cos ( \frac{1}{(x-3)^2+(y-4)^2}) \leq \lim_{(x,y) \rightarrow (3,4)} (x^2-9) $

or, $ 0 \leq \lim_{(x,y) \rightarrow (3,4)} (x^2-9) \cos ( \frac{1}{(x-3)^2+(y-4)^2}) \leq 0 $ .

Hence , $ \lim_{(x,y) \rightarrow (3,4)} (x^2-9) \cos ( \frac{1}{(x-3)^2+(y-4)^2}) =0 $

Am I right ? Any help ?

MAS
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  • More succinctly, $$\left|\cos\left(\frac{1}{(x-3)^2+(y-4)^2}\right)\right|\le 1$$Hence, $$0\le \left|(x^2-9)\cos\left(\frac{1}{(x-3)^2+(y-4)^2}\right)\right|\le |x^2-9|$$ – Mark Viola Jul 07 '17 at 03:21
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    Your general approach is entirely correct, but You'll need to be careful about the claim $9-x^2\le f(x, y)\le x^2 - 9$ as this inequality is only true for $x<3$. You'll probably want to use $|f(x, y)|\le |x^2 - 9|$ instead. – Kajelad Jul 07 '17 at 03:21
  • The answer is $ 0$ . Right ? – MAS Jul 07 '17 at 03:22

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