Why Did Rudin Use Absolute Value When Calculating the Derivative?

[Bachelier]

On Page 106 in baby rudin (diff. chapter) when he tries to calculate the derivative of the fuction


$$f(x) = \begin{cases}
x^2 sin(\frac{1}{x}) & \textrm{ if }x ≠ 0 \\
0 & \textrm{ if }x = 0 \\
\end{cases}$$

rudin used the absolute value in trying to compute the limit as $t → 0$

$i.e$

$\left|\frac{f(t) - f(0)}{t - 0}\right| = \left|t \ sin(\frac{1}{x})\right| ≤ |t|$

Why the abs. value?

 

[Bacle2]

I think it has to see with the fact that sin(-x)=-sin(x) . The limit then will be non-negative. Since is negative
in the 4th quadrant.

 

[Useful nucleus]

Absolute convergence implies convergence as in chapter 3 of the same book. And it turns out in this example checking absolute convergence is more obvious.

 

[lavinia]

It just shows that the Newton quotient at zero is squeezed by a number that goes to zero with t.

 

[Bachelier]

Thank you all