Question

1. $lim_{x

ightarrow4}\frac{sqrt{5x}-2sqrt{5}}{x - 4}$
$f(x)=$
$c=$

1. $lim_{x

ightarrow7}$
$f(x)=$
$c=$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction $\frac{\sqrt{5x}-2\sqrt{5}}{x - 4}$ by $\frac{\sqrt{5x}+2\sqrt{5}}{\sqrt{5x}+2\sqrt{5}}$. We get $\frac{(\sqrt{5x}-2\sqrt{5})(\sqrt{5x}+2\sqrt{5})}{(x - 4)(\sqrt{5x}+2\sqrt{5})}$. Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $5x-20$. So the fraction is $\frac{5x - 20}{(x - 4)(\sqrt{5x}+2\sqrt{5})}$.

Step2: Simplify the fraction

Factor out 5 from the numerator: $\frac{5(x - 4)}{(x - 4)(\sqrt{5x}+2\sqrt{5})}$. Cancel out the common factor $(x - 4)$ (since $x\to4$ but $x
eq4$), we have $\frac{5}{\sqrt{5x}+2\sqrt{5}}$.

Step3: Evaluate the limit

Substitute $x = 4$ into $\frac{5}{\sqrt{5x}+2\sqrt{5}}$. We get $\frac{5}{\sqrt{5\times4}+2\sqrt{5}}=\frac{5}{2\sqrt{5}+2\sqrt{5}}=\frac{5}{4\sqrt{5}}=\frac{\sqrt{5}}{4}$.

Answer:

$\frac{\sqrt{5}}{4}$