Consider the following function : $$f(x) = \begin{cases} x^2-9 \ & \text{if}\ x \le 4 \\[2ex] \frac{2x^2-9x+4}{x-4} & \text{if}\ x >4 \end{cases}$$
Now I want to show that this function is continuous at $x =4$ using $\epsilon-\delta$ definition. Seeing this at first I thought it might be easy to show but the calculation becomes messy and I can't find a proper $\delta$. I will show here some of my calculation,
Let $\epsilon >0$ now we need $\delta > 0$ so that $|x-4| <\delta \Rightarrow |f(x)-f(4)| < \epsilon.$ We can form two cases for limit from both sides. First I check for $x \le4$.
We need $|f(x)-f(4)| < \epsilon \Rightarrow |x^2-16| <\epsilon \Rightarrow 16-\epsilon < x^2 < 16+\epsilon \Rightarrow x\in (-\sqrt{16+\epsilon}, -\sqrt{16-\epsilon}).$
Also, we need $|x-4| < \delta \Rightarrow 4-\delta < x < 4+\delta$. So how do I choose $\delta$ now?? I know how to do this but I can't figure here. I am getting confused. Any hints?
Let $\epsilon >0$ and assume that the $\delta$ that we are going to choose is less than $1$
Note that if $|x-4|<\delta$ and $\delta <1$ then we have $3<x<5$
Thus for $x< 4$, $$|f(x)-7|=|x^2-9-7|=$$
$$|x^2-16|=|x+4||x-4|<9|x-4|<9\delta$$
On the other hand for $x\ge 4$, $$ |f(x)-7|=|\frac {2x^2-5x+4}{x-4}-7|=$$
$$|2x-1-7|=|2x-8|=2|x-4|<2\delta$$ Thus if we choose $$\delta <\min \{\epsilon /9,1\} $$ then we have $$|f(x)-7|<\epsilon$$
Thus the function is continuous at $x=4$