lugreget9

Answered

2022-01-14

The proof that if f(x) is continuous at $x=c$ , then $\frac{1}{f\left(x\right)}$ is continuous at $x=c$

If f(x) is continuous at$x=c,\text{}f\left(x\right)\ne 0\text{}\text{then}\text{}\frac{1}{f\left(x\right)}$ is continuous at $x=c$ .

My proof: for any$\u03f5>0$ , we need to show there exists $\delta >0$ , such that whenever $|x-c|<\delta$ , we have $|\frac{1}{f\left(x\right)}-\frac{1}{f\left(c\right)}|<\u03f5$ .

$|\frac{1}{f\left(x\right)}-\frac{1}{f\left(c\right)}|=\left|\frac{f\left(x\right)-f\left(c\right)}{f\left(x\right)f\left(c\right)}\right|$

$=|f\left(x\right)-f\left(c\right)|\frac{1}{\left|f\left(x\right)f\left(c\right)\right|}$

$=|f\left(x\right)-f\left(c\right)|\frac{1}{\left|f\left(x\right)\right|}\frac{1}{\left|f\left(c\right)\right|}$

Since f is continuous at c, thus$|f\left(x\right)-f\left(c\right)|<\u03f5$ . I need to find an upper bound of $\left|f\left(x\right)\right|$ .

Because$|f\left(x\right)-f\left(c\right)|<\u03f5$ , we then have

$\left|f\left(c\right)\right|=|f\left(c\right)-f\left(x\right)+f\left(x\right)|\le |f\left(c\right)-f\left(x\right)|+\left|f\left(x\right)\right|$ .

$<\u03f5+\left|f\left(x\right)\right|$

Thus, we find$\left|f\left(x\right)\right|>\left|f\left(c\right)\right|-\u03f5$ . Equivalently, $\frac{1}{\left|f\left(x\right)\right|}<\frac{1}{\left|f\left(c\right)\right|-\u03f5}$ . (Not rigorous here, since $\left|f\left(c\right)\right|-\u03f5$ could be negative, how to remedy?)

Finally, (ignore not rigorous part)

If f(x) is continuous at

My proof: for any

Since f is continuous at c, thus

Because

Thus, we find

Finally, (ignore not rigorous part)