Prove using the ε-δ definition that f(x) is continuous at the following points?

Question

How can I use the $\epsilon$-$\delta$ definition to do this for $f(x) = \frac{x + 2}{x^2 + x + 1}$ at $x=1$ and for $f(x) = \log(x^2 + 1)$ at $x=0$? I can do it for a normal function but I get confused when it gets to the stage of $|f(x)-f(a)|<\epsilon$ and I have to restrict this in terms of $|x-0|<\delta$ or $|x-1|<\delta$. Thanks

Answer 1

I'm assuming that by normal function you mean a polynomial. In the cases you've given, the problems work very similarly, but it is often less clear what choices to make for $\delta$.

To begin, fix $\epsilon>0$. At some point, we will need a $\delta$, but we don't know what it should be yet, so we'll just assume it is greater than 0. Suppose we have $|x-1|<\delta$. Then \begin{align} |f(x)-f(1)|&=\left|\frac{x+2}{x^2+x+1}-\frac{1+2}{1^2+1+1}\right|\\ &=\left|\frac{x+2}{x^2+x+1}-1\right|\\ &=\left|\frac{1-x^2}{x^2+x+1}\right|\\ &=\frac{|1-x||1+x|}{|x^2+x+1|}\\ &\leq\frac{\delta|x+1|}{|x^2+x+1|} \end{align}

At this point, we can determine what we want $\delta$ to be. Setting $\delta=\min\{1,\frac{\epsilon}{3}\}$ is enough in this case, see if you can figure out how to go from there.

Answer 2

For the second function, recalling that $f(0)=0$ and that $f(x) \geq 0$ since $1+x^2\geq 1$, you pick $\varepsilon>0$ and start with the inequality $$ |f(x)-f(0)| <\varepsilon, $$ namely $$ \log(1+x^2)<\varepsilon. $$ Since log is a strictly increasing function, your last inequality is equivalent to $$ 1+x^2 < e^\varepsilon. $$ Hence $$ -\sqrt{e^\varepsilon-1} < x < \sqrt{e^\varepsilon-1}, $$ and you can define $\delta=\sqrt{e^\varepsilon-1}>0$.