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!
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!
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 $a=0$.
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 $|x|+|\sin x| <\delta$.