The Proof Is In The Pudding... Maybe

I'm working through an interesting book on techniques for deciphering and writing mathematical proofs. Though I suck at lots of math (and non-math) things, I'm not actually alone in sucking at reading/writing proofs. One reason is they are almost completely deemphasized in any average math class you take before college (and I include geometry, even though you do write proofs in that course—the style taught therein isn't adequate preparation for what you encounter later). Pre-college math is more focused on answers to contrived problems, and I'm not going to make any statement as to whether that is "right" or "wrong"—it is what it is, but eventually you hit a wall if you can't "handle" proofs.

So here's my first crack at writing a simple one. Also I got to figure out how to display math in a browser, which is a nice bonus. Hooray for MathJax!

Note: I disabled MathJax's image-font fallback, since that feature takes up like 120 MB of space. Hopefully as long as you have a relatively new browser, this shouldn't matter, but let me know if the proper math symbols don't seem to be displaying.

Suppose that $$\lim_{x\to c} f(x) = L$$ for finite $$L$$ and that $$k$$ is constant. Then: $\lim_{x\to c} k \, f(x) = k L$

Proof:

• Given $$\lim_{x\to c} f(x) = L$$ we recognize that for all $$\varepsilon > 0$$, there exists $$\delta > 0$$ such that:
• $\left|\, f(x)-L \,\right|<\varepsilon \quad\text{when}\quad 0<\left|\, x-c \,\right|<\delta$
• Assume $$k \neq 0$$ and define $$\varepsilon_2 = \frac{\varepsilon}{|\,k\,|}$$.
• Then there likewise exists $$\delta_2 > 0$$ such that $$0<\left| \, x-c \, \right|<\delta_2$$ implies $$\left| \, f(x)-L \, \right|<\varepsilon_2$$.
• $\left|\, f(x)-L \,\right|<\frac{\varepsilon}{|\,k\,|}$
• $= \left|\,k \left( f(x)-L \right) \, \right|<\varepsilon$
• $= \left|\,k\, f(x) - kL \,\right|<\varepsilon$
• Thus, $$\lim_{x\to c} k \, f(x) = k L$$ for $$k \neq 0$$.
• Now assume $$k = 0$$: $$\quad \lim_{x\to c} 0 = 0$$.
• Therefore, $$\lim_{x\to c} k \, f(x) = k L$$ for all constant $$k$$.