If $\left| x-2\right| < \frac{1}{100}$ Show That $\left| x^2 -4 \right| < \frac{1}{10}$

Question

I am working through a problem involving absolute value inequalities, and I just can't seem to get on the right track. The problem is, if:

$$\left| x-2 \right| < \frac{1}{100} \text{,} $$

show that

$$ \left| x^2-4 \right| < \frac{1}{10}$$

must be true. I thought it might have something to do with $x^2-4$ being a difference of squares with $x-2$ as a factor, but that just lead me in circles. Any pointers or hints would be appreciated! Thank you.

Answer 1

If $|x-2|<\frac{1}{100}$, then $|x| < 2+ \frac {1}{100} <8$

So

$$|x^2-4| = |x-2||x+2| < \frac{1}{100} (|x|+2)<\frac{1}{10}.$$

Answer 2

Consider $f(x) = x^2 - 4$. $|x-2|<\frac{1}{100}$ means $f$ is within $\frac{1}{100}$ of $2$. In particular, we want to consider $f(x)$ for $x \in (1.99, 2.01)$. $f$ is increasing and continuous in this interval, and hence the image set is $(1.99^2 - 4, 2.01^2 - 4) = (\frac{-399}{10000}, \frac{401}{10000}) \subset (-\frac{1}{10}, \frac{1}{10})$.

It is easy to see, therefore, that $|f| < \frac {1}{10}$ in this interval.