$\epsilon - \delta$ definition to prove that f is a continuous function.

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?

• It might be just simpler to prove that for any pair of continuous functions $f,g$, the product $fg$ is also continuous, and to prove that if $g(x)\neq 0$ then $1/g$ is continuous at $x$. Dec 10 '13 at 2:19
• if i prove it this way, do you have any suggestions? Dec 10 '13 at 2:20
• You can find proof in Spivak's or Apostol's calculus, probably in Rudin's Principles too. Dec 10 '13 at 2:28
• Dec 10 '13 at 2:29

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 $\delta=\min(a,b).$