To prove that [tex]\(\lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+a}-\sqrt{x})=\frac{a}{2}\)[/tex], let's go through the calculations step by step.
First, we start with the expression we want to find the limit of:
[tex]\[
\lim_{x \to \infty} \sqrt{x}(\sqrt{x + a} - \sqrt{x})
\][/tex]
To simplify, we multiply both the numerator and the denominator by the conjugate of the expression inside the limit:
[tex]\[
\text{Conjugate} = \sqrt{x + a} + \sqrt{x}
\][/tex]
So, multiplying by the conjugate, we have:
[tex]\[
\lim_{x \to \infty} \sqrt{x}(\sqrt{x + a} - \sqrt{x}) \cdot \frac{\sqrt{x + a} + \sqrt{x}}{\sqrt{x + a} + \sqrt{x}}
\][/tex]
This can be broken down into:
[tex]\[
\lim_{x \to \infty} \frac{\sqrt{x} \left( \sqrt{x + a} - \sqrt{x} \right) \left( \sqrt{x + a} + \sqrt{x} \right)}{\sqrt{x + a} + \sqrt{x}}
\][/tex]
Notice that in the numerator, we have a difference of squares:
[tex]\[
(\sqrt{x + a} - \sqrt{x})(\sqrt{x + a} + \sqrt{x}) = (x + a - x) = a
\][/tex]
Therefore, our limit expression now simplifies to:
[tex]\[
\lim_{x \to \infty} \frac{\sqrt{x} \cdot a}{\sqrt{x + a} + \sqrt{x}}
\][/tex]
Factoring [tex]\(\sqrt{x}\)[/tex] out of the denominator:
[tex]\[
\lim_{x \to \infty} \frac{a \sqrt{x}}{\sqrt{x}(\sqrt{1 + \frac{a}{x}} + 1)}
\][/tex]
Simplify further:
[tex]\[
\lim_{x \to \infty} \frac{a \sqrt{x}}{\sqrt{x} \left( \sqrt{1 + \frac{a}{x}} + 1 \right)}
\][/tex]
Since [tex]\(\sqrt{x}\)[/tex] is in both the numerator and the denominator, it cancels out:
[tex]\[
\lim_{x \to \infty} \frac{a}{\sqrt{1 + \frac{a}{x}} + 1}
\][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{a}{x}\)[/tex] approaches 0. So, the denominator simplifies to:
[tex]\[
\sqrt{1 + 0} + 1 = 1 + 1 = 2
\][/tex]
Thus, the limit simplifies to:
[tex]\[
\frac{a}{2}
\][/tex]
Therefore, we have successfully shown that:
[tex]\[
\lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+a}-\sqrt{x}) = \frac{a}{2}
\][/tex]
