Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers 2

up vote 7 down vote accepted

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 $a=0$.

share|improve this answer
add comment

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 $|x|+|\sin x| <\delta$.

share|improve this answer
1  
It might be helpful to note $|sinx|\leq|x|\;\;\;\forall x\in R$. –  bobobinks Jan 30 '11 at 1:55
    
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! –  fdart17 Jan 30 '11 at 2:45
1  
@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 $y$. –  Jonas Meyer Jan 30 '11 at 2:54
    
Yes that's a typo sorry thanks. –  fdart17 Jan 30 '11 at 2:57
1  
@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 $\sqrt{x^2+\sin^2x}<\delta$. –  Jonas Meyer Jan 30 '11 at 23:15
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.