Question

1. (limlimits_{x \to a} \frac{x - a}{sqrt{x} - sqrt{a}}, a > 0)

Explanation:

Step1: Rationalize the denominator

We can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x} + \sqrt{a}$.
So we have:
\[

$$\begin{align*} \lim_{x ightarrow a}\frac{x - a}{\sqrt{x}-\sqrt{a}}&=\lim_{x ightarrow a}\frac{(x - a)(\sqrt{x}+\sqrt{a})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}\\ \end{align*}$$

\]

Step2: Simplify the denominator

Using the difference of squares formula $(m - n)(m + n)=m^{2}-n^{2}$, the denominator $(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})$ simplifies to $x - a$.
So the expression becomes:
\[
\lim_{x
ightarrow a}\frac{(x - a)(\sqrt{x}+\sqrt{a})}{x - a}
\]

Step3: Cancel out common terms

We can cancel out the common factor $(x - a)$ from the numerator and the denominator (since $x
ightarrow a$ but $x
eq a$ at the limit point, so we can cancel).
After canceling, we get:
\[
\lim_{x
ightarrow a}(\sqrt{x}+\sqrt{a})
\]

Step4: Substitute the limit value

Now we substitute $x = a$ into the expression $\sqrt{x}+\sqrt{a}$.
We have $\sqrt{a}+\sqrt{a}=2\sqrt{a}$.

Answer:

$2\sqrt{a}$