Proof of Continuity of $f(x) = x^2 + x - 1$

Question

Am supposed to show proof that $f(x) = x^2 + x - 1$ is continuous for all real numbers a. Doing the general Delta-Epsilon proof, I have and understand how to prove $x^2$ alone. In particular I am having trouble finding a proper $δ$ in terms of $ε$.

$|f(x) - f(a)| = |(x^2+x-1)-(a^2+a-1)| = |(x^2+x)-(a^2+a)| = |(x^2-a^2)+(x-a)|$

From here, I can say $|(x^2-a^2)+(x-a)|\leq |x^2-a^2|+|x-a|$ by Triangle Inequality, so then letting $δ < 1$ (as the proof of $x^2$ alone goes), I get $x < a+1$, so then $|x^2-a^2+x-a| < |2a+1+x-a|\leq |2a+1|+|x-a|$ again by Triangle Inequality. $|2a+1|+|x-a|$ can be $<δ$ if I let $δ$ be the $min(1,-|2a+1|+ε)$, then $δ$ is not necessarily positive.

I know you can also say that $|(x^2-a^2)+(x-a)| = |x-a||x+a|+|x-a|$, but I have not been able to think of a $δ$ such that $|x-a||x+a|+|x-a|<ε$.

I believe my troubles are coming from not knowing how to properly arrange an equivalent form of $|f(x) - f(a)|$, but after spending a couple of hours at it, I haven't made any progress. Could anyone advise?

EDIT: Thank you everyone for giving me the summation of continuous functions route, but we have not yet proven that, and cannot use it. I realize this is a quick and easy proof, but the point of the exercise was to not use it, I should have specified it.

Answer 1

Basically, we want to factor out the $|x - a|$, then use the fact that $\delta < 1$ in order to bound what's left over.

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Given any $\epsilon > 0$, let $\delta = \min(1, \tfrac{\epsilon}{1 + |2a + 1|}) > 0$. Then if $0 < |x - a| < \delta$, observe that: \begin{align*} |(x^2 + x - 1) - (a^2 + a - 1)| &= |(x^2 - a^2) + (x - a)| \\ &= |(x - a)(x + a) + (x - a)(1)| \\ &= |x - a||x + a + 1| \\ &= |x - a||(x - a) + (2a + 1)| \\ &\leq |x - a|(|x - a| + |2a + 1|) &\text{by the triangle inequality}\\ &< |x - a|(1 + |2a + 1|) &\text{since } |x - a| < \delta \leq 1 \\ &< \tfrac{\epsilon}{1 + |2a + 1|}(1 + |2a + 1|) &\text{since } |x - a| < \delta \leq \tfrac{\epsilon}{1 + |2a + 1|} \\ &= \epsilon \end{align*} as desired.

Answer 2

The sum of continuous functions is continuous. This is not too difficult to prove. Let $f$ and $g$ be continuous functions. Then $$\lim_{x\rightarrow x_0}(f+g)(x) = \lim_{x\rightarrow x_0}f(x) + \lim_{x\rightarrow x_0}g(x) = f(x_0) + g(x_0) = (f+g)(x_0)$$ Above, I used the theorem that limits split across a sum.

Now we can separately prove $x^2$ is continuous, prove $x$ is continuous, and prove $-1$ is continuous. Then use the fact that the sum of continuous functions is continuous, and you're done.