Answer:
lim = 1/6
Step-by-step explanation:
Apparently, we have ...
[tex]g(t)=\sqrt{t+7}[/tex]
It often works well to multiply by the conjugate of a difference that approaches zero.
[tex]\displaystyle\lim_{h\to 0}{\frac{g(2+h)-g(2)}{h}}=\lim_{h\to 0}{\frac{\sqrt{(2+h)+7}-\sqrt{2+7}}{h}}\\\\=\lim_{h\to 0}\frac{(\sqrt{9+h}-\sqrt{9})(\sqrt{9+h}+\sqrt{9})}{h(\sqrt{9+h}+\sqrt{9})}=\lim_{h\to 0}{\frac{(9+h)-9}{h(\sqrt{9+h}+\sqrt{9})}}\\\\=\lim_{h\to0}{\frac{h}{h(\sqrt{9+h}+3)}}=\left.\frac{1}{\sqrt{9+h}+3}\right|_{h=0}=\boxed{\frac{1}{6}}[/tex]
