$$
f(x) = x^2 + 3x \text{ at x = } 1
$$
|
$$
f(x) = 3 -5x^2 \text{ at x = } 1
$$
|
$$
f(x) = -\frac{1}{x^2} \text{ at x = } 2
$$
|
$$
f(x) = 2\sqrt{x} \text{ at x = } 1
$$
$f(x)=2\sqrt{x} \text{ at x = }1$
$\lim\limits_{h \rightarrow 0}\frac{f(x_{0}+h)-f(x_{0})}{h}
\\
\lim\limits_{h \rightarrow 0}\frac{2\sqrt{(1+h)}-2\sqrt{(1)}}{h}$
$\lim\limits_{h \rightarrow 0}\frac{2\sqrt{(1+h)}-2}{h}$
$\lim\limits_{h \rightarrow 0}\frac{2\sqrt{(1+h)}-2}{h}\Bigg(\frac{2\sqrt{(1+h)}+2}{2\sqrt{(1+h)}+2}\Bigg)$
$\lim\limits_{h \rightarrow 0}\frac{(2\sqrt{(1+h)}-2)(2\sqrt{(1+h)}+2)}{h(2\sqrt{(1+h)}+2)}$
$\lim\limits_{h \rightarrow 0}\frac{4(1+h)-4}{h(2\sqrt{(1+h)}+2)}$
$\lim\limits_{h \rightarrow 0}\frac{4h}{h(2\sqrt{(1+h)}+2)}
\\
\lim\limits_{h \rightarrow 0}\frac{4}{(2\sqrt{(1+h)}+2)} =
\\
\frac{4}{(2\sqrt{(1+0)}+2)} =
\frac{4}{(2+2)} = \frac{4}{(4)} = 1$
|