Answer
$y=\dfrac{x}{4}+\dfrac{5}{4}$
Work Step by Step
Given: $f(x)=\sqrt{x+1}$; $(3,2)$
Here, we have $m=\lim\limits_{h\to 0}\dfrac{f(3+h)-f(3)}{h}=\lim\limits_{h\to 0}\dfrac{\sqrt{h+4}-2}{h}$
$m=\lim\limits_{h\to 0} \dfrac{h+4-4}{h(\sqrt{h+4}+2)}$
This gives:
or, $m=\lim\limits_{h\to 0} \dfrac{1}{\sqrt{h+4}+2)}=\dfrac{1}{4}$
Next, the equation of the tangent line is given as:
$y-2=m(x-3)$
This implies that
$y-2=\dfrac{1}{4}(x-3)$
Thus, $y=\dfrac{x}{4}+\dfrac{5}{4}$
