Show that $g(x)=\sqrt{x^{2}+4}$ is continuous at $x=1$ Let $g(x)=\sqrt{x^{2}+4}$ from $\mathbb{R}\rightarrow \mathbb{R}$. I want to show that $g$ is continuous at $x=1$.
I have to show that for any $\epsilon>0$, there exists a $\delta>0$ such that $|x-1|< \delta \implies |g(x)-g(1)|<\epsilon$.
So I do some rough work first and start with $$|g(x)-g(1)|$$ $$=|\sqrt{x^{2}+4} -\sqrt{5}|$$ $$\leq|\sqrt{x^{2}+4}| + |\sqrt{5}|$$.
I am stuck on what to do after. The plan is to choose the right $\delta$. 

As suggested in a comment of imranfat, multiply by the conjugate, i.e.
\begin{align*}
\sqrt{x^2+4}-\sqrt{5}=\frac{(\sqrt{x^2+4}-\sqrt{5})(\sqrt{x^2+4}+\sqrt{5})}{\sqrt{x^2+4}+\sqrt{5}}=\frac{x^2-1}{\sqrt{x^2+4}+\sqrt{5}}=\frac{(x-1)(x+1)}{\sqrt{x^2+4}+\sqrt{5}}.
\end{align*}
Now, if $|x-1|<\delta$, then $|x+1|=|x-1+2|\leq|x-1|+2\leq\delta+2$. Since $\sqrt{x^2+4}+\sqrt{5}\geq\sqrt{5}\geq1$, it follows that
\begin{align*}
|\sqrt{x^2+4}-\sqrt{5}|=\frac{|x-1||x+1|}{\sqrt{x^2+4}+\sqrt{5}}\leq\delta(\delta+2).
\end{align*}
Can you go on from here?
