Exercise 5.4.5

Question

Show that
$$
\frac{g\left(x_{m}\right)-g(0)}{x_{m}-0}=m+1
$$
and use this to prove that $g^{\prime}(0)$ does not exist.

Ulisse Mini · Accepted Answer

Note that $h_n(x_m) = x_m$ for $0 \leq n \leq m$ and $h_n(x_m) = 0$ for $n > m$, therefore $g(x_m) = (m + 1) x_m$. Clearly $g(0) = 0$ , so
$$\frac{g\left(x_{m}\right)-g(0)}{x_{m}-0}=m+1$$

If $g'(0)$ existed, then
$$\lim_{x \to 0} \frac{g(x)-g(0)}{x-0}$$
would be well defined. But we've just identified a sequence $x_m$ approaching 0 for which this expression grows without bound, and hence this limit cannot exist, and therefore $g'(0)$ does not exist.

Exercise 5.4.5

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Exercise 5.4.5

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