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Let $f: U\subseteq\mathbb{R^2}\to\mathbb{R}$, $(x_0,y_0)\in U$ and suppose that $\lim\limits_{(x,y)\to(x_0,y_0)} f(x,y) = 0$. Prove that $$\lim\limits_{(x,y)\to(x_0,y_0)} \frac{\sin f(x,y)}{f(x,y)} = 1 $$

Any hint is welcome because I'm not even sure how to begin. I think it is not necessary to use the delta-epsilon definition if we reduce it to the case of a single variable limit. It is possible? Or this doesn't strictly proof the double limit?

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3 Answers 3

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Hint: for small $t$

$$ \sin t \sim t. $$

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You’ll probably want to to apply the following proposition. Note that it does not depend on $f$ and $g$ being continuous.

Prop. Suppose $\lim_{x\to p} f(x) = c$ and $\lim_{y\to c} g(y) = L$. Then $\lim_{x \to p} g(f(x)) = L$.

Proof. Our goal is to make $|g(f(x))-L|$ arbitrarily small. So fix $\epsilon>0$. Suppose that $|g(y) - L| < \epsilon$ for all $y$ satisfying $|y-c|<\delta$. Such a number $\delta$ exists by the fact that $g(y) \to L$. Suppose, further, that $|f(x)-c|<\delta$ for all $x$ satisfying $|x-p|<\delta'$. Such a number $\delta'$ exists by the fact that $f(x) \to c$. Then, for all such $x$, $|g(f(x))-L|<\epsilon$. So, indeed, $g(f(x)) \to L$.

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  • $\begingroup$ It doesn't depend on f and g being continuous? Here's a counterexample: $f(t)=g(t)=0$ for $t \neq 0$ and $f(0)=g(0)=1$. You need to assume $g$ is continuous at $c$. $\endgroup$
    – JasonJones
    Oct 15, 2020 at 21:57
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Remember, limit of a function is the function of the limit. I.E. $$\lim_{x\rightarrow a} f(x)=h$$ $$\lim_{x\rightarrow a} = g(f(x)) = g \left(\lim_{x\rightarrow a} f(x) \right) = g(h)$$ Then apply the other hint about $\sin(x) \approx x$ for small $x$. You could also use L'Hopital's rule.

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    $\begingroup$ Be careful, this holds only for $g$ continuous :) $\endgroup$
    – Ivo Terek
    Apr 18, 2015 at 1:39
  • $\begingroup$ Of course, this is only true if the function is continuous at the limit point. $\endgroup$
    – MPW
    Apr 18, 2015 at 1:40

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