27. find $lim_{h \to 0}\frac{f(x + h)-f(x)}{h}$, if $f(x)=sqrt{5x}$. show all work!!

Explanation:

Step1: Substitute function into the limit

First, find $f(x + h)$ where $f(x)=\sqrt{5x}$, so $f(x + h)=\sqrt{5(x + h)}$. Then the limit becomes $\lim_{h\rightarrow0}\frac{\sqrt{5(x + h)}-\sqrt{5x}}{h}$.

Step2: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{5(x + h)}+\sqrt{5x}}{\sqrt{5(x + h)}+\sqrt{5x}}$. We get $\lim_{h\rightarrow0}\frac{(\sqrt{5(x + h)}-\sqrt{5x})(\sqrt{5(x + h)}+\sqrt{5x})}{h(\sqrt{5(x + h)}+\sqrt{5x})}$. Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator is $5(x + h)-5x$.

Step3: Simplify the numerator

Expand $5(x + h)-5x$ to get $5x+5h - 5x = 5h$. So the limit is now $\lim_{h\rightarrow0}\frac{5h}{h(\sqrt{5(x + h)}+\sqrt{5x})}$.

Step4: Cancel out the common factor

Cancel out the $h$ terms in the numerator and denominator. We have $\lim_{h\rightarrow0}\frac{5}{\sqrt{5(x + h)}+\sqrt{5x}}$.

Step5: Evaluate the limit

As $h\rightarrow0$, substitute $h = 0$ into the expression. We get $\frac{5}{\sqrt{5x}+\sqrt{5x}}=\frac{5}{2\sqrt{5x}}=\frac{\sqrt{5}}{2\sqrt{x}}$.

Answer:

$\frac{\sqrt{5}}{2\sqrt{x}}$