Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$,  $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\frac {3h}{6+h} - 10}{h} $$ I am stuck at this part

$$\lim_{h\to 0}\frac{4\sqrt{4+h}-8}{h}+\lim_{h\to 0}\frac{2e^{\tfrac{3h}{6+h}}-2}{h}$$$$=4\lim_{h\to 0}\frac{\sqrt{4+h}-\sqrt4}{h}+2\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{h}$$Rationalising the first limit and adjusting the denominator of the second limit, $$ =4\lim_{h\to 0}\frac{h}{h\cdot(\sqrt{4+h}+\sqrt4)}+2\times 3\lim_{h\to0}\frac{e^{\tfrac{3h}{6+h}}-1}{\frac{3h}{6+h}\cdot (6+h)}$$ Using direct substitution of $h=0$ on the first limit and the formula $\lim \limits_{U\to 0}\dfrac{e^U-1}{U}=1$ on the second limit,$$=4\cdot\frac{1}{2+2}+\frac{2\times3}{6\times 1}=2.$$
