Answer
$$\frac{1}{2 \sqrt{5}}$$
$$y=\frac{1}{2 \sqrt{5}} x+\frac{9}{2 \sqrt{5}}$$
Work Step by Step
Let $f(x)=\sqrt{x+4}$. Then
\begin{aligned}
f^{\prime}(1) &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{h+5}-\sqrt{5}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{h+5}-\sqrt{5}}{h} \cdot \frac{\sqrt{h+5}+\sqrt{5}}{\sqrt{h+5}+\sqrt{5}} \\
&=\lim _{h \rightarrow 0} \frac{h}{h(\sqrt{h+5}+\sqrt{5})}\\
&=\lim _{h \rightarrow 0} \frac{1}{\sqrt{h+5}+\sqrt{5}}=\frac{1}{2 \sqrt{5}}
\end{aligned}
The tangent line at $a=1$ is
\begin{aligned}
y&=f^{\prime}(1)(x-1)+f(1)\\
&=\frac{1}{2 \sqrt{5}}(x-1)+\sqrt{5}\\
&=\frac{1}{2 \sqrt{5}} x+\frac{9}{2 \sqrt{5}}
\end{aligned}
