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Evaluate the limit:
[tex]\[ \lim_{x \rightarrow \infty} \left( \sqrt{x + a} - \sqrt{x} \right) \][/tex]


Sagot :

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]