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

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.

• $x^2-4$ is not a perfect square, it factors as $(x-2)(x+2)$. – Zain Patel Jul 10 '16 at 14:53
• Difference of squares* – Logan Toll Jul 10 '16 at 14:54

If $|x-2|<\frac{1}{100}$, then $|x| < 2+ \frac {1}{100} <8$
$$|x^2-4| = |x-2||x+2| < \frac{1}{100} (|x|+2)<\frac{1}{10}.$$
• There's two things I'm still confused about: Why is the $8$ important? Also I don't follow the last step; how do we know that $\frac{1}{100}\left(\left| x\right|+2\right) <\frac{1}{10}$? – Logan Toll Jul 10 '16 at 15:44
• @LoganToll : I just make that $8$ so that at the end $|x|+2 <8 +2 = 10$. – user99914 Jul 10 '16 at 15:57
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.