The limit of $f(x) = \frac{1}{x^2 + 5x - 24}$ at $x=4$

Question

I'm working through Advanced Calculus: Theory and Practice by John S. Petrovic and is currently working on problem 3.4.2, which is as follows:

Find the limit and prove that the result is correct using the definition (epsilon-delta) of the limit: $$ \lim_{x \to 4} \frac{1}{x^2 + 5x - 24} $$

This is my attempt: $$ \lim_{x \to 4} \frac{1}{x^2 + 5x - 24} = \frac{1}{12} $$

We need $| x - 4 | < \delta$, and an expression for $\delta$ dependent on $\epsilon$, therefore we focus on \begin{align} \Biggl|f(x) - \frac{1}{12}\Biggr| &= \Biggl|\frac{12 - x^2 - 5x + 24}{12(x^2 + 5x - 24)}\Biggr| \\ &= \Biggl|\frac{-(x + 9)(x - 4)}{12(x + 8)(x - 3)}\Biggr| \\ &= \Biggl|\frac{(x + 9)(x - 4)}{12(x + 8)(x - 3)}\Biggr| \\ &\leq \frac{1}{12} \frac{|x+9|\,|x-4|}{|x+8|\,|x-3|}. \end{align}

(It is the next step where I am pretty sure I'm using the inequalities incorrectly.)

Now let $\delta_1 = 1$, which implies that $-1 < x - 4 < 1 \implies 3 < x < 5$ and therefore:

1. $12 < x + 9 < 14 \implies |x + 9| < 14$
2. $11 < x + 8 < 13 \implies |x + 8| > 11$
3. $0 < x - 3 < 2 \implies |x - 3| > 0$

Therefore, $$ \frac{1}{12} \frac{|x+9|\,|x-4|}{|x+8|\,|x-3|} < \frac{1}{12}\frac{14\,|x - 4|}{11 \cdot 0} \cdots $$

As you can see the inequality $|x - 3| > 0$ renders the expression undefined. Where am I going wrong? If you supply me with a complete solution, please do not use any "fancy" theorems as this problem is presented at the beginning of the continuity chapter and therefore should be solvable using only simple methods.

Thank you.

Answer 1

You're doing everything correctly, including finding upper bounds for linear factors in the numerator and lower bounds for linear factors in the denominator.

The problem is that the initial choice of $\delta_1 = 1$ is too big since the limit $x \to 4$ is too close to $x = 3$, where the factor $x - 3$ vanishes. Although the arithmetic is bit messier, if you begin with $\delta_1 = \tfrac12$, and follow the same steps, everything should work out.

In general, if you're writing a proof for a limit as $x \to a$ for a rational function with roots $r_1, \dots, r_n$ in the denominator, you need to choose an initial $$ \delta_1 < \min \bigl\{ |r_1 - a|, \dots, |r_n - a| \bigr\}. $$

Answer 2

Graphing on desmos, there's no hole, sharp point, or asymptote there, so it's continuous there.