Question

find the limit. (limlimits_{h \to 0^+} \frac{sqrt{h^2 + 2h + 3} - sqrt{3}}{h}) (limlimits_{h \to 0^+} \frac{sqrt{h^2 + 2h + 3} - sqrt{3}}{h} = square)

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{h^2 + 2h + 3}+\sqrt{3}$.

$$\begin{align*} &\lim_{h ightarrow 0^{+}}\frac{\sqrt{h^2 + 2h + 3}-\sqrt{3}}{h}\times\frac{\sqrt{h^2 + 2h + 3}+\sqrt{3}}{\sqrt{h^2 + 2h + 3}+\sqrt{3}}\\ =&\lim_{h ightarrow 0^{+}}\frac{(h^2 + 2h + 3)-3}{h(\sqrt{h^2 + 2h + 3}+\sqrt{3})}\\ \end{align*}$$

Step2: Simplify the numerator

Simplify the numerator: $(h^2 + 2h + 3)-3 = h^2 + 2h$.

$$ \lim_{h ightarrow 0^{+}}\frac{h^2 + 2h}{h(\sqrt{h^2 + 2h + 3}+\sqrt{3})} $$

Step3: Cancel out the common factor

Cancel out the common factor $h$ from the numerator and the denominator (since $h
ightarrow 0^{+}$, $h
eq0$).

$$ \lim_{h ightarrow 0^{+}}\frac{h + 2}{\sqrt{h^2 + 2h + 3}+\sqrt{3}} $$

Step4: Substitute $h = 0$

Substitute $h = 0$ into the simplified expression.

$$ \frac{0 + 2}{\sqrt{0^2 + 2\times0 + 3}+\sqrt{3}}=\frac{2}{\sqrt{3}+\sqrt{3}}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} $$

Answer:

$\frac{\sqrt{3}}{3}$