Question

Find the derivative of $99x$ at $x=100$

Answer 1

Let $f\left(x\right)=99x$
Thus $f^{{}^{\prime }}\left(100\right)=\underset{h\to 0}{lim}\frac{f(100+h)-f\left(100\right)}{h}$
$=\underset{h\to 0}{lim}\frac{99(100+h)-99\left(100\right)}{h}$
$=\underset{h\to 0}{lim}\frac{99\times 100+99h-99\times 100}{h}$
$=\underset{h\to 0}{lim}\frac{99h}{h}=99$