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If $f$ is continuous at $(0,0)$, is $g(x) := f(x, \sin x)$ continuous at $0$?

Intuitively this seems true because as you approach 0 the $\sin$ function also approaches 0 from both sides. Any push on how to prove this would be appreciated!

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You can also use sequences: If $(x_n)$ is a sequence with $x_n\to0$ then also $\sin x_n\to0$, so the continuity of $f$ at $(0,0)$ implies that $g(x_n) = f(x_n,\sin x_n) \to f(0,0) = g(0)$.

This is a special case of a more general theorem: If $h$ is continuous at $a$ and $f$ is continuous at $h(a)$, then $g = f\circ h$ is continuous at $a$. In the above example one has $h(x) = (x,\sin x)$ and $a=0$.

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Since $f$ is continuous at $(0,0)$ you know that for all $\varepsilon\gt 0$ there exists $\delta\gt 0$ such that $|x|+|\sin x|\lt\delta$ implies $|f(x,\sin x) - f(0,0)|\lt\varepsilon$. Since $|f(x,\sin x)-f(0,0)|=|g(x)-g(0)|$, it will be enough to show that there is a $\delta'\gt 0$ such that $|x|\lt\delta'$ implies $|x|+|\sin x| <\delta$.

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    $\begingroup$ It might be helpful to note $|sinx|\leq|x|\;\;\;\forall x\in R$. $\endgroup$ – Benji Jan 30 '11 at 1:55
  • $\begingroup$ Your first sentence, this is from the limit definition of multi variable funtions? Wouldn't it be $\sqrt{x^2 + sinx} < \delta$ ? Sorry I'm trying to understand where that's coming from! $\endgroup$ – fdart17 Jan 30 '11 at 2:45
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    $\begingroup$ @Fdart17: Do you mean $\sqrt{x^2+\sin^2 x}$? These are 2 equivalent ways to state continuity for multivariable functions, and they come from 2 equivalent ways to measure distances between points. If you prefer to use the Euclidean distance, feel free. If you like working with sequences, $((x_n,y_n))_n^\infty$ converges to $(x,y)$ if and only if $(x_n)$ converges to $x$ and $(y_n)$ converges to $y$. $\endgroup$ – Jonas Meyer Jan 30 '11 at 2:54
  • $\begingroup$ Yes that's a typo sorry thanks. $\endgroup$ – fdart17 Jan 30 '11 at 2:57
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    $\begingroup$ @Fdart17: It is just a special case. $f$ is continuous at $(0,0)$ means that for all $\varepsilon\gt0$ there is a $\delta\gt0$ such that $\sqrt{x^2+y^2}<\delta$ implies $|f(x,y)-f(0,0)|<\varepsilon$. This is true for all $(x,y)$ satisfying the condition, so in particular it holds if $y=\sin(x)$ and $\sqrt{x^2+\sin^2x}<\delta$. $\endgroup$ – Jonas Meyer Jan 30 '11 at 23:15

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