Answer
$$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}=\frac{2\sqrt5}{5}$$
Work Step by Step
$$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}$$
To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal.
$$A=\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}$$
Here we need to multiply both nominator and denominator by $\sqrt{h^2+4h+5}+\sqrt5$ to eliminate the $h$ in the denominator.
- Nominator: $(\sqrt{h^2+4h+5}-\sqrt5)(\sqrt{h^2+4h+5}+\sqrt5)=h^2+4h+5-5=h^2+4h$
- Denominator: $h(\sqrt{h^2+4h+5}+\sqrt5)$
Hence, $$A=\lim_{h\to0^+}\frac{h^2+4h}{h(\sqrt{h^2+4h+5}+\sqrt5)}$$ $$A=\lim_{h\to0^+}\frac{h+4}{\sqrt{h^2+4h+5}+\sqrt5}$$ $$A=\frac{0+4}{\sqrt{0^2+4\times0+5}+\sqrt5}$$ $$A=\frac{4}{\sqrt5+\sqrt5}=\frac{4}{2\sqrt5}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}$$
Therefore, $$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt5}{h}=\frac{2\sqrt5}{5}$$
