To find the limit [tex]\(\lim_{x \to \infty} (\sqrt{x + a} - \sqrt{x})\)[/tex], let's take a detailed, step-by-step approach.
1. Initial expression:
[tex]\(\lim_{x \to \infty} (\sqrt{x + a} - \sqrt{x})\)[/tex]
2. Rationalize the expression:
To simplify the expression, we rationalize it by multiplying and dividing by the conjugate. The conjugate of [tex]\(\sqrt{x + a} - \sqrt{x}\)[/tex] is [tex]\(\sqrt{x + a} + \sqrt{x}\)[/tex].
[tex]\[
\lim_{x \to \infty} \frac{(\sqrt{x + a} - \sqrt{x})(\sqrt{x + a} + \sqrt{x})}{\sqrt{x + a} + \sqrt{x}}
\][/tex]
3. Simplify the numerator:
Using the difference of squares formula [tex]\((\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B\)[/tex], we simplify the numerator:
[tex]\[
(\sqrt{x + a})^2 - (\sqrt{x})^2 = (x + a) - x = a
\][/tex]
Thus, the expression becomes:
[tex]\[
\lim_{x \to \infty} \frac{a}{\sqrt{x + a} + \sqrt{x}}
\][/tex]
4. Simplify the denominator:
As [tex]\(x\)[/tex] approaches infinity, [tex]\(x + a\)[/tex] is approximately [tex]\(x\)[/tex], hence [tex]\(\sqrt{x + a}\)[/tex] is approximately [tex]\(\sqrt{x}\)[/tex]. This simplification allows us to approximate the denominator.
[tex]\[
\sqrt{x + a} + \sqrt{x} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x}
\][/tex]
5. Revise the limit expression:
[tex]\[
\lim_{x \to \infty} \frac{a}{2\sqrt{x}}
\][/tex]
6. Evaluate the limit:
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\sqrt{x}\)[/tex] also approaches infinity. Therefore, the fraction [tex]\(\frac{a}{2\sqrt{x}}\)[/tex] approaches zero because the numerator [tex]\(a\)[/tex] is a constant and the denominator [tex]\(2\sqrt{x}\)[/tex] grows without bound.
[tex]\[
\lim_{x \to \infty} \frac{a}{2\sqrt{x}} = 0
\][/tex]
Thus, we find that:
[tex]\[
\lim_{x \to \infty} (\sqrt{x + a} - \sqrt{x}) = 0
\][/tex]
