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Use the $\epsilon$-$\delta$ definition of continuity to prove that $$f(x)= \frac{x+1}{x-4}$$ is continuous at every point $c \in R \backslash \{4\}$.

My attempt is

let $c \in R \backslash \{4\}$.

Given $\epsilon > 0$, we choose $\delta =$ something I do not know. Then | x-c | $< \delta$ implies that$$|f(x) – f(c)| = \Big|\frac{x+1}{x-4} – \frac{c+1}{c-4}\Big| = \Big|\frac{-5|x-c|}{|x-4||c-4|}\Big| $$ (this is the part I do not know how to simplify to get to something $= \epsilon$.

Could someone please check my direction for the proof? If it is right, could you help me with the parts that I said I do not know?

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When $x$ is near $4$ the function $f(x)=\frac{x+1}{x-4}$ gets large. So if a fixed $c \neq 4$ is chosen at which we want to show $f$ is continuous, we need the $\delta$ to do two jobs: first it needs to keep $x$ near enough to $c$ that it avoids being near $4$. This can be done by assuming $x$ is closer to $c$ than it is to $4$, which is done provided $|x-c|<|c-4|/2.$ Now we have from the triangle inequality that

$$|c-x|+|x-4| \ge |c-4|. \\ |x-4| \ge |c-4|-|x-c|.$$

From this and the already assumed $|x-c|<|c-4|/2$ it follows that $|x-4|>|c-4|/2,$ which on taking reciprocals is

$$\frac{1}{|x-4|} < \frac{2}{|c-4|}.\tag{1}$$

Now also we have

$$|f(x)-f(c)|=\frac{5|x-c|}{|c-4||x-4|}<\frac{10|x-c|}{|c-4|^2}, \tag{2}$$

where at the second inequality we used $(1).$

Now if $|x-c|<(1/10)|c-4|^2 \cdot \varepsilon$ the right side of $(2)$ will be less than $\varepsilon$.

So our $\delta$ must be both at most $|4-c|/2$ and $(1/10)|c-4|^2 \varepsilon$, in order to guarantee $|f(x)-f(c)|< \varepsilon.$ Usually when two inequalities $\delta<a,\delta<b$ have been imposed to make something small, one writes $\delta=\min(a,b).$

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