Find the derivative of 99x at x = 100.

Let f(x) = 99x. Accordingly,

$f^{\prime}(100)=\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h}$

$=\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h}$

$=\lim _{h \rightarrow 0} \frac{99 \times 100+99 h-99 \times 100}{h}$

$=\lim _{h \rightarrow 0} \frac{99 h}{h}$

$=\lim _{h \rightarrow 0}(99)=99$

Thus, the derivative of $99 x$ at $x=100$ is 99 .