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To determine the limit of the expression [tex]\(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\)[/tex] as [tex]\(x\)[/tex] approaches 3, let's proceed step-by-step:
1. Substitute [tex]\(x = 3\)[/tex] into the expression:
[tex]\[ f(x) = \frac{x-3}{\sqrt{x-2} - \sqrt{4-x}} \][/tex]
Plugging in [tex]\(x = 3\)[/tex]:
[tex]\[ f(3) = \frac{3-3}{\sqrt{3-2} - \sqrt{4-3}} = \frac{0}{\sqrt{1} - \sqrt{1}} = \frac{0}{0} \][/tex]
This is an indeterminate form, so we need to use algebraic techniques to resolve it.
2. Simplify the denominator by rationalizing it. To do this, we'll multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \text{Conjugate of } (\sqrt{x-2} - \sqrt{4-x}) \text{ is } (\sqrt{x-2} + \sqrt{4-x}) \][/tex]
[tex]\[ \text{Thus, multiply } \frac{x-3}{\sqrt{x-2} - \sqrt{4-x}} \text{ by } \frac{\sqrt{x-2} + \sqrt{4-x}}{\sqrt{x-2} + \sqrt{4-x}} \][/tex]
3. Perform the multiplication:
[tex]\[ \frac{x-3}{\sqrt{x-2} - \sqrt{4-x}} \cdot \frac{\sqrt{x-2} + \sqrt{4-x}}{\sqrt{x-2} + \sqrt{4-x}} = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{(\sqrt{x-2})^2 - (\sqrt{4-x})^2} \][/tex]
Simplify the denominator using the difference of squares:
[tex]\[ = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{(x-2) - (4-x)} = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{x-2-4+x} = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{2x-6} \][/tex]
Note that [tex]\(2x - 6 = 2(x-3)\)[/tex]:
[tex]\[ = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{2(x-3)} \][/tex]
4. Cancel out the common factor [tex]\((x-3)\)[/tex] from the numerator and the denominator:
[tex]\[ = \frac{\sqrt{x-2} + \sqrt{4-x}}{2} \][/tex]
5. Evaluate the limit of the simplified expression as [tex]\(x\)[/tex] approaches 3:
[tex]\[ \lim_{x \to 3} \frac{\sqrt{x-2} + \sqrt{4-x}}{2} \][/tex]
Substitute [tex]\(x = 3\)[/tex]:
[tex]\[ = \frac{\sqrt{3-2} + \sqrt{4-3}}{2} = \frac{\sqrt{1} + \sqrt{1}}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1 \][/tex]
Therefore, the limit of [tex]\(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\)[/tex] as [tex]\(x\)[/tex] approaches 3 is [tex]\(\boxed{1}\)[/tex].
1. Substitute [tex]\(x = 3\)[/tex] into the expression:
[tex]\[ f(x) = \frac{x-3}{\sqrt{x-2} - \sqrt{4-x}} \][/tex]
Plugging in [tex]\(x = 3\)[/tex]:
[tex]\[ f(3) = \frac{3-3}{\sqrt{3-2} - \sqrt{4-3}} = \frac{0}{\sqrt{1} - \sqrt{1}} = \frac{0}{0} \][/tex]
This is an indeterminate form, so we need to use algebraic techniques to resolve it.
2. Simplify the denominator by rationalizing it. To do this, we'll multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \text{Conjugate of } (\sqrt{x-2} - \sqrt{4-x}) \text{ is } (\sqrt{x-2} + \sqrt{4-x}) \][/tex]
[tex]\[ \text{Thus, multiply } \frac{x-3}{\sqrt{x-2} - \sqrt{4-x}} \text{ by } \frac{\sqrt{x-2} + \sqrt{4-x}}{\sqrt{x-2} + \sqrt{4-x}} \][/tex]
3. Perform the multiplication:
[tex]\[ \frac{x-3}{\sqrt{x-2} - \sqrt{4-x}} \cdot \frac{\sqrt{x-2} + \sqrt{4-x}}{\sqrt{x-2} + \sqrt{4-x}} = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{(\sqrt{x-2})^2 - (\sqrt{4-x})^2} \][/tex]
Simplify the denominator using the difference of squares:
[tex]\[ = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{(x-2) - (4-x)} = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{x-2-4+x} = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{2x-6} \][/tex]
Note that [tex]\(2x - 6 = 2(x-3)\)[/tex]:
[tex]\[ = \frac{(x-3)(\sqrt{x-2} + \sqrt{4-x})}{2(x-3)} \][/tex]
4. Cancel out the common factor [tex]\((x-3)\)[/tex] from the numerator and the denominator:
[tex]\[ = \frac{\sqrt{x-2} + \sqrt{4-x}}{2} \][/tex]
5. Evaluate the limit of the simplified expression as [tex]\(x\)[/tex] approaches 3:
[tex]\[ \lim_{x \to 3} \frac{\sqrt{x-2} + \sqrt{4-x}}{2} \][/tex]
Substitute [tex]\(x = 3\)[/tex]:
[tex]\[ = \frac{\sqrt{3-2} + \sqrt{4-3}}{2} = \frac{\sqrt{1} + \sqrt{1}}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1 \][/tex]
Therefore, the limit of [tex]\(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\)[/tex] as [tex]\(x\)[/tex] approaches 3 is [tex]\(\boxed{1}\)[/tex].
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