Answer :

To evaluate the limit [tex]\(\lim _{x \rightarrow \infty}(\sqrt{x-a}-\sqrt{x-b})\)[/tex], we can simplify the expression by using some algebraic manipulation techniques.

1. Rewrite the Expression Using the Conjugate:
Start by multiplying and dividing the given expression by the conjugate of the terms inside the limit. The conjugate of [tex]\(\sqrt{x-a} - \sqrt{x-b}\)[/tex] is [tex]\(\sqrt{x-a} + \sqrt{x-b}\)[/tex].

[tex]\[ \sqrt{x-a} - \sqrt{x-b} = \frac{(\sqrt{x-a} - \sqrt{x-b})(\sqrt{x-a} + \sqrt{x-b})}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]

When we multiply the terms inside the numerator:

[tex]\[ (\sqrt{x-a} - \sqrt{x-b})(\sqrt{x-a} + \sqrt{x-b}) = (x-a) - (x-b) = x - a - x + b = b - a \][/tex]

2. Simplified Form:
The rewritten expression is:

[tex]\[ \sqrt{x-a} - \sqrt{x-b} = \frac{b-a}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]

3. Analyze the Denominator as [tex]\( x \to \infty \)[/tex]:
As [tex]\( x \)[/tex] approaches infinity, both [tex]\(\sqrt{x-a}\)[/tex] and [tex]\(\sqrt{x-b}\)[/tex] will approach [tex]\(\sqrt{x}\)[/tex]. This is because the subtraction of constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex] becomes negligible for very large [tex]\( x \)[/tex].

Thus, the expression in the denominator [tex]\(\sqrt{x-a} + \sqrt{x-b}\)[/tex] can be approximated to:

[tex]\[ \sqrt{x-a} + \sqrt{x-b} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x} \][/tex]

4. Simplify the Limit:
Substituting this back into the original expression gives:

[tex]\[ \lim _{x \rightarrow \infty} \frac{b-a}{\sqrt{x-a} + \sqrt{x-b}} \approx \lim _{x \rightarrow \infty} \frac{b-a}{2\sqrt{x}} \][/tex]

5. Evaluate the Limit:
Now, as [tex]\( x \)[/tex] approaches infinity, [tex]\( \sqrt{x} \)[/tex] also approaches infinity. Therefore, the fraction [tex]\(\frac{b-a}{2\sqrt{x}}\)[/tex]:

[tex]\[ \frac{b-a}{2\sqrt{x}} \to \frac{b-a}{\infty} = 0 \][/tex]

6. Conclusion:
The limit of [tex]\( \sqrt{x-a} - \sqrt{x-b} \)[/tex] as [tex]\( x \)[/tex] approaches infinity is:

[tex]\[ \lim _{x \rightarrow \infty}(\sqrt{x-a}-\sqrt{x-b}) = 0 \][/tex]

Therefore, the result is:
[tex]\[ 0 \][/tex]