Question

find the limit.
lim_{h
ightarrow0}\frac{sqrt{h^{2}+12h + 5}-sqrt{5}}{h}
lim_{h
ightarrow0}\frac{sqrt{h^{2}+12h + 5}-sqrt{5}}{h}=square

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{h^{2}+12h + 5}+\sqrt{5}}{\sqrt{h^{2}+12h + 5}+\sqrt{5}}$. We get $\lim_{h
ightarrow0}\frac{(\sqrt{h^{2}+12h + 5}-\sqrt{5})(\sqrt{h^{2}+12h + 5}+\sqrt{5})}{h(\sqrt{h^{2}+12h + 5}+\sqrt{5})}$.
Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(h^{2}+12h + 5)-5=h^{2}+12h$. So the limit is $\lim_{h
ightarrow0}\frac{h^{2}+12h}{h(\sqrt{h^{2}+12h + 5}+\sqrt{5})}$.

Step2: Simplify the fraction

Cancel out the common factor $h$ in the numerator and denominator. We have $\lim_{h
ightarrow0}\frac{h(h + 12)}{h(\sqrt{h^{2}+12h + 5}+\sqrt{5})}=\lim_{h
ightarrow0}\frac{h + 12}{\sqrt{h^{2}+12h + 5}+\sqrt{5}}$.

Step3: Evaluate the limit

Substitute $h = 0$ into the simplified function. $\frac{0 + 12}{\sqrt{0^{2}+12\times0+5}+\sqrt{5}}=\frac{12}{2\sqrt{5}}=\frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}$.

Answer:

$\frac{6\sqrt{5}}{5}$