If you look at 1/g(x) - 1/g(c) = (g(c) - g(x)) / (g(c) \* g(x)), you can make the numerator small using continuity of g at c. So as long as you can get a positive lower bound on the denominator, you get the continuity you need. But since |g(c)| > 0 and g is continuous at c, you can get a choice of delta such that |g(x) - g(c)| < |g(c)|/2, and then |g(x)| > |g(c)|/2.

If you look at 1/g(x) - 1/g(c) = (g(c) - g(x)) / (g(c) \* g(x)), you can make the numerator small using continuity of g at c. So as long as you can get a positive lower bound on the denominator, you get the continuity you need. But since |g(c)| &gt; 0 and g is continuous at c, you can get a choice of delta such that |g(x) - g(c)| &lt; |g(c)|/2, and then |g(x)| &gt; |g(c)|/2.

Let D be a subset of R, and let f,g : D –&gt; R be functions which are continuous at c in D. Use the epsilon-delta definition of continuity to prove that f/g is continuous at c assuming g(c) =/= 0.

I'm having trouble setting it up and going about the proof. I know I'm supposed to use the triangle inequality but I can't get that part right I think. Currently, I have |(f(x)/g(x)) - (f(c)/(g(c))| = |(f(x)/(g(x)) - (f(c)/(g(x)) + (f(c)/(g(x)) - (f(c)/(g(c))| &lt;= |1/g(x)||f(x) - f(c)| + |f(c)||1/(g(x)) - 1/(g(c))| but I don't know if this is right. 

[Real Analysis] Function continuity proof