IDNLearn.com: Your go-to resource for finding expert answers. Discover reliable and timely information on any topic from our network of knowledgeable professionals.
Sagot :
To find the limit [tex]\(\lim_{x \rightarrow \infty}(\sqrt{x-a}-\sqrt{x-b})\)[/tex], let's start by exploring the behavior of the function as [tex]\(x\)[/tex] approaches infinity.
1. Writing the expression in a more convenient form:
Begin with the original expression:
[tex]\[ \sqrt{x-a} - \sqrt{x-b} \][/tex]
2. Rationalize the expression:
To simplify this, we'll multiply and divide by the conjugate of the expression (i.e., [tex]\(\sqrt{x-a} + \sqrt{x-b}\)[/tex]). This helps us remove the square roots from the numerator:
[tex]\[ \left(\sqrt{x-a} - \sqrt{x-b}\right) \times \frac{\sqrt{x-a} + \sqrt{x-b}}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]
Simplify the result:
[tex]\[ \frac{(\sqrt{x-a} - \sqrt{x-b})(\sqrt{x-a} + \sqrt{x-b})}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]
3. Simplify the numerator:
The numerator now becomes a difference of squares:
[tex]\[ (\sqrt{x-a} - \sqrt{x-b})(\sqrt{x-a} + \sqrt{x-b}) = (x-a) - (x-b) = -a + b = b-a \][/tex]
So, we simplify our expression to:
[tex]\[ \frac{b-a}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]
4. Evaluate the denominator:
As [tex]\(x\)[/tex] approaches infinity, both [tex]\(\sqrt{x-a}\)[/tex] and [tex]\(\sqrt{x-b}\)[/tex] will behave similarly to [tex]\(\sqrt{x}\)[/tex], since [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants. Therefore:
[tex]\[ \sqrt{x-a} + \sqrt{x-b} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x} \][/tex]
5. Substitute back and further simplify:
Plug the approximate values back into the expression:
[tex]\[ \frac{b-a}{2\sqrt{x}} \][/tex]
6. Evaluate the limit:
Now, consider the limit as [tex]\(x\)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{b-a}{2\sqrt{x}} \][/tex]
Since [tex]\(\sqrt{x}\)[/tex] increases without bound as [tex]\(x\)[/tex] goes to infinity, the fraction [tex]\(\frac{b-a}{2\sqrt{x}}\)[/tex] will approach 0. Thus, we have:
[tex]\[ \lim_{x \rightarrow \infty}(\sqrt{x-a}-\sqrt{x-b}) = 0 \][/tex]
So, the desired limit is:
[tex]\[ \boxed{0} \][/tex]
1. Writing the expression in a more convenient form:
Begin with the original expression:
[tex]\[ \sqrt{x-a} - \sqrt{x-b} \][/tex]
2. Rationalize the expression:
To simplify this, we'll multiply and divide by the conjugate of the expression (i.e., [tex]\(\sqrt{x-a} + \sqrt{x-b}\)[/tex]). This helps us remove the square roots from the numerator:
[tex]\[ \left(\sqrt{x-a} - \sqrt{x-b}\right) \times \frac{\sqrt{x-a} + \sqrt{x-b}}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]
Simplify the result:
[tex]\[ \frac{(\sqrt{x-a} - \sqrt{x-b})(\sqrt{x-a} + \sqrt{x-b})}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]
3. Simplify the numerator:
The numerator now becomes a difference of squares:
[tex]\[ (\sqrt{x-a} - \sqrt{x-b})(\sqrt{x-a} + \sqrt{x-b}) = (x-a) - (x-b) = -a + b = b-a \][/tex]
So, we simplify our expression to:
[tex]\[ \frac{b-a}{\sqrt{x-a} + \sqrt{x-b}} \][/tex]
4. Evaluate the denominator:
As [tex]\(x\)[/tex] approaches infinity, both [tex]\(\sqrt{x-a}\)[/tex] and [tex]\(\sqrt{x-b}\)[/tex] will behave similarly to [tex]\(\sqrt{x}\)[/tex], since [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants. Therefore:
[tex]\[ \sqrt{x-a} + \sqrt{x-b} \approx \sqrt{x} + \sqrt{x} = 2\sqrt{x} \][/tex]
5. Substitute back and further simplify:
Plug the approximate values back into the expression:
[tex]\[ \frac{b-a}{2\sqrt{x}} \][/tex]
6. Evaluate the limit:
Now, consider the limit as [tex]\(x\)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{b-a}{2\sqrt{x}} \][/tex]
Since [tex]\(\sqrt{x}\)[/tex] increases without bound as [tex]\(x\)[/tex] goes to infinity, the fraction [tex]\(\frac{b-a}{2\sqrt{x}}\)[/tex] will approach 0. Thus, we have:
[tex]\[ \lim_{x \rightarrow \infty}(\sqrt{x-a}-\sqrt{x-b}) = 0 \][/tex]
So, the desired limit is:
[tex]\[ \boxed{0} \][/tex]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.
