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I am trying to prove continuity of the function

$$\text{Let } f(x)=\begin{cases} f_1(x) = \frac{1}{x} & \text{if }x\in (0,\frac{1}{2}] \\ f_2(x) = 2 & \text{if }x\in(\frac{1}{2},1) \end{cases}$$

I have proved the continuity of $f_1$ and $f_2$ over their respective intervals but I am now stuck at the left continuity of $f$ at point $c=\frac12$.

So far I have, where for $x<\frac12$ for all $\epsilon>0$ there exists a $\delta>0$ such that $\vert x-\frac12\vert<\delta$ implies:

$$\vert f(x)-f(\frac12)\vert=\vert\frac1x-2\vert$$ from here I am pretty sure this is wrong but I tried doing: $$\vert\frac1x-2\vert=\vert\frac{2}{2x}-\frac{4x}{2x}\vert=\frac{2}{x}\vert\frac12-x\vert=\frac{2}{x}\vert x-\frac12\vert$$ and if this somehow worked, I don't really know what I should choose my $\delta$ to be.

Thanks in advance.

DoubleV
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1 Answers1

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This is like most $\delta$-$\epsilon$ proofs with nonlinear functions. You need to make a preliminary stipulation/supposition to give an upper bound on $1/x$. (You also want to write your proof so that given $\epsilon>0$, you will provide a $\delta>0$ that works.)

So we have $$\left|\frac1x-2\right| = \left|\frac{1-2x}x\right| = \frac 2x\left|\frac12-x\right|.$$ Let's now stipulate that $\dfrac 38<x<\dfrac12$, so that $\dfrac 1x<\dfrac83$. This will mean that $$\left|\frac1x-2\right|=\frac 2x\left|\frac12-x\right|<2\cdot \frac83\left|\frac12-x\right|.$$ Let's now set $\delta = \min\left(\frac18,\frac{3\epsilon}{16}\right)$. I leave it to you to finish the argument that if $0<\frac12-x<\delta$, then $\left|\frac1x-2\right|<\epsilon$. Just turn the scratchwork above into a one-line final estimate.

Ted Shifrin
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