Answer :
To solve the limit [tex]\(\lim_{x \to \infty} (\sqrt{x + a} - \sqrt{x})\)[/tex], let's go through the process step-by-step.
First, observe the form of the given expression:
[tex]\[ \sqrt{x + a} - \sqrt{x} \][/tex]
As [tex]\(x\)[/tex] approaches infinity, both [tex]\(\sqrt{x + a}\)[/tex] and [tex]\(\sqrt{x}\)[/tex] also approach infinity. Therefore, we have an indeterminate form of type [tex]\(\infty - \infty\)[/tex]. To resolve this, we will rationalize the expression by multiplying and dividing by the conjugate of the expression [tex]\(\sqrt{x + a} - \sqrt{x}\)[/tex]. The conjugate is [tex]\(\sqrt{x + a} + \sqrt{x}\)[/tex].
Now, let’s proceed with the rationalization:
[tex]\[ \sqrt{x + a} - \sqrt{x} = \frac{(\sqrt{x + a} - \sqrt{x})(\sqrt{x + a} + \sqrt{x})}{\sqrt{x + a} + \sqrt{x}} \][/tex]
Using the difference of squares formula, the numerator becomes:
[tex]\[ (\sqrt{x + a})^2 - (\sqrt{x})^2 = (x + a) - x = a \][/tex]
Thus, the expression is transformed into:
[tex]\[ \sqrt{x + a} - \sqrt{x} = \frac{a}{\sqrt{x + a} + \sqrt{x}} \][/tex]
Now, we need to find the limit as [tex]\(x\)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{a}{\sqrt{x + a} + \sqrt{x}} \][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\sqrt{x + a}\)[/tex] and [tex]\(\sqrt{x}\)[/tex] both behave similarly because [tex]\(a\)[/tex] is a constant and does not affect the leading behavior as [tex]\(x\)[/tex] grows very large. Therefore:
[tex]\[ \lim_{x \to \infty} \left(\sqrt{x + a}\right) = \lim_{x \to \infty} \left(\sqrt{x}\right) \][/tex]
So the expression in the denominator becomes approximately:
[tex]\[ \sqrt{x + a} + \sqrt{x} \approx 2\sqrt{x} \][/tex]
Substituting this into our limit gives:
[tex]\[ \lim_{x \to \infty} \frac{a}{\sqrt{x+a} + \sqrt{x}} \approx \lim_{x \to \infty} \frac{a}{2\sqrt{x}} \][/tex]
Now, simplifying further:
[tex]\[ \lim_{x \to \infty} \frac{a}{2\sqrt{x}} \][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\sqrt{x}\)[/tex] also approaches infinity, making the whole fraction approach zero:
[tex]\[ \lim_{x \to \infty} \frac{a}{2\sqrt{x}} = 0 \][/tex]
Thus, the limit is:
[tex]\[ \lim_{x \to \infty} (\sqrt{x + a} - \sqrt{x}) = 0 \][/tex]
First, observe the form of the given expression:
[tex]\[ \sqrt{x + a} - \sqrt{x} \][/tex]
As [tex]\(x\)[/tex] approaches infinity, both [tex]\(\sqrt{x + a}\)[/tex] and [tex]\(\sqrt{x}\)[/tex] also approach infinity. Therefore, we have an indeterminate form of type [tex]\(\infty - \infty\)[/tex]. To resolve this, we will rationalize the expression by multiplying and dividing by the conjugate of the expression [tex]\(\sqrt{x + a} - \sqrt{x}\)[/tex]. The conjugate is [tex]\(\sqrt{x + a} + \sqrt{x}\)[/tex].
Now, let’s proceed with the rationalization:
[tex]\[ \sqrt{x + a} - \sqrt{x} = \frac{(\sqrt{x + a} - \sqrt{x})(\sqrt{x + a} + \sqrt{x})}{\sqrt{x + a} + \sqrt{x}} \][/tex]
Using the difference of squares formula, the numerator becomes:
[tex]\[ (\sqrt{x + a})^2 - (\sqrt{x})^2 = (x + a) - x = a \][/tex]
Thus, the expression is transformed into:
[tex]\[ \sqrt{x + a} - \sqrt{x} = \frac{a}{\sqrt{x + a} + \sqrt{x}} \][/tex]
Now, we need to find the limit as [tex]\(x\)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{a}{\sqrt{x + a} + \sqrt{x}} \][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\sqrt{x + a}\)[/tex] and [tex]\(\sqrt{x}\)[/tex] both behave similarly because [tex]\(a\)[/tex] is a constant and does not affect the leading behavior as [tex]\(x\)[/tex] grows very large. Therefore:
[tex]\[ \lim_{x \to \infty} \left(\sqrt{x + a}\right) = \lim_{x \to \infty} \left(\sqrt{x}\right) \][/tex]
So the expression in the denominator becomes approximately:
[tex]\[ \sqrt{x + a} + \sqrt{x} \approx 2\sqrt{x} \][/tex]
Substituting this into our limit gives:
[tex]\[ \lim_{x \to \infty} \frac{a}{\sqrt{x+a} + \sqrt{x}} \approx \lim_{x \to \infty} \frac{a}{2\sqrt{x}} \][/tex]
Now, simplifying further:
[tex]\[ \lim_{x \to \infty} \frac{a}{2\sqrt{x}} \][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\sqrt{x}\)[/tex] also approaches infinity, making the whole fraction approach zero:
[tex]\[ \lim_{x \to \infty} \frac{a}{2\sqrt{x}} = 0 \][/tex]
Thus, the limit is:
[tex]\[ \lim_{x \to \infty} (\sqrt{x + a} - \sqrt{x}) = 0 \][/tex]