Let $f(x)= (x² + 1) / (√x²-1)$. Determine the domain of f and prove that f is continuous for every point in its domain.
So domain would be determined by what makes $(√x²−1) ≠ 0$? This would be all of $x∈R$ except $x≠1$ or $x≠ -1$. Then you would have to prove continuity for the intervals $(-∞,-1), (-1, 1),$ and $(1, ∞)$. And $(x²+1)>0$ simply by its nature. And $(√x²-1)>0$ again because a square root cannot be negative. Thus the domain of $f$ is $(1, ∞)$.
Proof has to be made using Epsilon-Delta definition: $∀ε>0 ∃δ>0$ such that if $x∈I$ and $|x−x₀|<δ$ then $|f(x)−f(x₀)|<ε$.
Let $g(x)=x²+1$. And let $h(x)=√(x²-1)$. If we can show that each of these is continuous for all $x\in (1,∞)$, then by the rules of continuity, we can show $f=g/h$ is continuous as well.
Let's start with $g$.
Proof: Let $g(x)=x²+1$.
Let $\epsilon>0$ be arbitrary. Let $x₀\in (1, ∞)$ be arbitrary.
Let $\delta g=min ❴1, \epsilon/(2|x₀|+1)❵$.
Let $x\in (1, ∞)$ and $|x-x₀|<\delta g$.
Thus $|g(x)-g(x₀)|=|x²+1-x₀²-1|=|x²-x₀²|<(2|x₀|+1)|x-x₀|<(2|x₀|+1)\delta g\leq(2|x₀|+1)(\epsilon/(2|x₀+1))=\epsilon$
Thus $|g(x)-g(x₀)|<\epsilon$. Therefore since $\epsilon$ is arbitrary, $g$ is continuous at $x₀$. And since $x₀$ is arbitrary, $g$ is continuous for all of $x\in (1, ∞)$. ∎
Now $h$.
Proof: Let $h(x)=√(x²-1)$.
Let $\epsilon>0$ be arbitrary. Let $x₀\in (1, ∞)$ be arbitrary.
Let $\delta h=?$.
Let $x\in (1, ∞)$ and $|x-x₀|<\delta h$.
Thus $|h(x)-h(x₀)|=|√(x²-1)-√(x₀²-1)|=?$ And this is where I get stuck.
The rest would go:
Thus $|h(x)-h(x₀)|<\epsilon$. Therefore since $\epsilon$ is arbitrary, $h$ is continuous at $x₀$. And since $x₀$ is arbitrary, $h$ is continuous for all of $x\in (1, ∞)$. ∎
Let $f(x)=g(x)/h(x)$.
Because $h(x)≠0$ for $x\in (1, ∞)$ then f is continuous for all $x\in (1, ∞)$∎
