To solve the limit [tex]\(\lim _{x \rightarrow \infty}(\sqrt{3x} - \sqrt{3x - 5})\)[/tex], follow these detailed steps:
1. Rationalize the expression:
The difference of two square roots can be handled by rationalizing to simplify the expression. To rationalize [tex]\(\sqrt{3x} - \sqrt{3x - 5}\)[/tex], multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3x} + \sqrt{3x - 5}\)[/tex]:
[tex]\[
\sqrt{3x} - \sqrt{3x - 5} = \frac{(\sqrt{3x} - \sqrt{3x - 5})(\sqrt{3x} + \sqrt{3x - 5})}{\sqrt{3x} + \sqrt{3x - 5}}
\][/tex]
2. Simplify the numerator:
The numerator becomes:
[tex]\[
(\sqrt{3x} - \sqrt{3x - 5})(\sqrt{3x} + \sqrt{3x - 5}) = (\sqrt{3x})^2 - (\sqrt{3x - 5})^2 = 3x - (3x - 5) = 5
\][/tex]
So the expression simplifies to:
[tex]\[
\frac{5}{\sqrt{3x} + \sqrt{3x - 5}}
\][/tex]
3. Examine the denominator as [tex]\(x\)[/tex] approaches infinity:
As [tex]\(x\)[/tex] approaches infinity, both [tex]\(\sqrt{3x}\)[/tex] and [tex]\(\sqrt{3x - 5}\)[/tex] grow large. We can factor out [tex]\(\sqrt{3x}\)[/tex] from the denominator:
[tex]\[
\sqrt{3x} + \sqrt{3x - 5} = \sqrt{3x} \left(1 + \sqrt{1 - \frac{5}{3x}}\right)
\][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{5}{3x}\)[/tex] approaches 0, making [tex]\(\sqrt{1 - \frac{5}{3x}}\)[/tex] approach [tex]\(\sqrt{1} = 1\)[/tex].
So the expression further simplifies to approximately:
[tex]\[
\frac{5}{\sqrt{3x}(1 + 1)} = \frac{5}{2\sqrt{3x}}
\][/tex]
4. Evaluate the limit:
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\sqrt{3x}\)[/tex] approaches infinity. Therefore:
[tex]\[
\frac{5}{2\sqrt{3x}} \rightarrow 0
\][/tex]
5. Conclusion:
Thus, the limit is:
[tex]\[
\lim _{x \rightarrow \infty} (\sqrt{3x} - \sqrt{3x - 5}) = 0
\][/tex]
Hence, the result is indeed [tex]\(0\)[/tex].
