Using epsilon-delta to prove $\lim_{x\to a}\frac{x+3}{x^2+5}$ exists.

Question

How can I formulate a rigorous epsilon-delta proof that $$\lim_{x\to a}\frac{x+3}{x^2+5}$$ exists for any a without the use of any helper theorems. I'm quite stumped and don't know where to begin.

Answer 1

In general, what you are aiming at ('without the use of any helper theorems') is not a good idea. These helper theorems (like $a_n\rightarrow a, b_n\rightarrow b \neq 0 \Rightarrow \frac{a_n}{b_n}\rightarrow \frac{a}{b}$ are extremely helpful to avoid unnecessary complicated calculations.

In the special case you are looking at the complexity is not that terrible. Just note that the only reasonable candidate for the limit is $\frac{a+3}{a^2+5}$. Then calculate the difference $$ \frac{(a+s)+3}{(a+s)^2+5}-\frac{a+3}{a^2+5}$$ for some real $s$ and write this as one fraction:

$$ \frac{((a+s)+3)(a^2+5)-(a+3)((a+s)^2+5)}{(a^2+5)[(a+s)^2+5]}$$

After a short calculation you will see that in the numerator many terms cancel out and what remains is just $$s(a^2+5) - (a+3)(2as+s^2)$$

This could be further simplified, but there is no need to do this. What is important here is that you have a factor of $s$ in each term. This means that if $|s|$ is small (which implies that $s^2$ is even smaller), the absolute value of this term smaller than $C|s|$ for some constant $C$ (which you could write down more explicitly in terms of $a$, but this is really not what a mathematician would do if this is not explicitly needed).

On the other hand the absolute value of the denominator is bounded from below by some constant $D>0$. So the whole expression is bounded, in the norm, by $\frac{C}{D}|s|$ for two constants $C,D$ which can be calculated and depend only on $a$. So if $\varepsilon >0$ is given you just need to choose $\delta>0$, such that $|s|<\delta$ implies $\frac{C}{D}|s| < \varepsilon$ (which I leave to you).