Using "epsilon-delta" notation prove that $f(x)=\frac{1}{x^2+4}$ is continuous at $x_0\in D(f)$.

Question

Using "epsilon-delta" notation prove that $f(x)=\frac{1}{x^2+4}$ is continuous in $x_0\in D(f)$.

Any ideas? I have a problem finding the right $\delta$.

Answer 1

$$\left|\frac{1}{x^2+4}-\frac{1}{x_0^4+4}\right|=\frac{|x+x_0|}{(x^2+4)(x_0^2+4)}|x-x_0|$$

Take $x\in [x_0-1,x_0+1]$ and set $$M=\max_{x\in[x_0-1,x_0+1]}\frac{|x+x_0|}{(x^2+4)(x_0^2+4)}$$ (why $M$ exist ?). Then, if $\varepsilon>0$, take $\delta=\min\{\frac{\varepsilon}{M},1\}$ to get $$\left|\frac{1}{x^2+4}-\frac{1}{x_0^4+4}\right|<\varepsilon$$ if $|x-x_0|<\delta$.