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To find the limit of the function [tex]\(\lim_{x \rightarrow 2}\left(\frac{\sqrt{x+2}-2}{\sqrt{x+7}-3}\right)\)[/tex], let's go through the steps in detail.
First, we need to understand the form of the limit as [tex]\(x\)[/tex] approaches 2. Substituting [tex]\(x = 2\)[/tex] directly into the function, we have:
[tex]\[ \frac{\sqrt{2+2}-2}{\sqrt{2+7}-3} = \frac{\sqrt{4}-2}{\sqrt{9}-3} = \frac{2-2}{3-3} = \frac{0}{0} \][/tex]
This indicates that we have an indeterminate form [tex]\(\frac{0}{0}\)[/tex], and we need to apply some algebraic manipulation or a limit theorem to resolve it.
One effective method to handle this situation is to rationalize the numerator and the denominator to eliminate the square roots. Let's start by rationalizing the numerator.
Consider the numerator [tex]\(\sqrt{x+2} - 2\)[/tex]. We multiply and divide by its conjugate:
[tex]\[ \sqrt{x+2} - 2 = \frac{(\sqrt{x+2} - 2)(\sqrt{x+2} + 2)}{\sqrt{x+2} + 2} = \frac{(x + 2) - 4}{\sqrt{x+2} + 2} = \frac{x - 2}{\sqrt{x+2} + 2} \][/tex]
Now, substitute this back into the original limit:
[tex]\[ \lim_{x \to 2} \frac{\sqrt{x+2} - 2}{\sqrt{x+7} - 3} = \lim_{x \to 2} \frac{\frac{x - 2}{\sqrt{x+2} + 2}}{\sqrt{x+7} - 3} \][/tex]
Next, handle the denominator [tex]\(\sqrt{x+7} - 3\)[/tex] by a similar rationalization:
[tex]\[ \sqrt{x+7} - 3 = \frac{(\sqrt{x+7} - 3)(\sqrt{x+7} + 3)}{\sqrt{x+7} + 3} = \frac{(x + 7) - 9}{\sqrt{x+7} + 3} = \frac{x - 2}{\sqrt{x+7} + 3} \][/tex]
Substitute this back into the modified expression:
[tex]\[ \frac{\frac{x - 2}{\sqrt{x+2} + 2}}{\frac{x - 2}{\sqrt{x+7} + 3}} = \lim_{x \to 2} \frac{x - 2}{\sqrt{x+2} + 2} \cdot \frac{\sqrt{x+7} + 3}{x - 2} \][/tex]
As the terms [tex]\(x - 2\)[/tex] in the numerator and denominator cancel out, we get:
[tex]\[ \lim_{x \to 2} \frac{\sqrt{x+7} + 3}{\sqrt{x+2} + 2} \][/tex]
Now we can substitute [tex]\(x = 2\)[/tex] directly into the simplified expression:
[tex]\[ \frac{\sqrt{2+7} + 3}{\sqrt{2+2} + 2} = \frac{\sqrt{9} + 3}{\sqrt{4} + 2} = \frac{3 + 3}{2 + 2} = \frac{6}{4} = \frac{3}{2} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 2}\left(\frac{\sqrt{x+2}-2}{\sqrt{x+7}-3}\right) = \frac{3}{2} \][/tex]
First, we need to understand the form of the limit as [tex]\(x\)[/tex] approaches 2. Substituting [tex]\(x = 2\)[/tex] directly into the function, we have:
[tex]\[ \frac{\sqrt{2+2}-2}{\sqrt{2+7}-3} = \frac{\sqrt{4}-2}{\sqrt{9}-3} = \frac{2-2}{3-3} = \frac{0}{0} \][/tex]
This indicates that we have an indeterminate form [tex]\(\frac{0}{0}\)[/tex], and we need to apply some algebraic manipulation or a limit theorem to resolve it.
One effective method to handle this situation is to rationalize the numerator and the denominator to eliminate the square roots. Let's start by rationalizing the numerator.
Consider the numerator [tex]\(\sqrt{x+2} - 2\)[/tex]. We multiply and divide by its conjugate:
[tex]\[ \sqrt{x+2} - 2 = \frac{(\sqrt{x+2} - 2)(\sqrt{x+2} + 2)}{\sqrt{x+2} + 2} = \frac{(x + 2) - 4}{\sqrt{x+2} + 2} = \frac{x - 2}{\sqrt{x+2} + 2} \][/tex]
Now, substitute this back into the original limit:
[tex]\[ \lim_{x \to 2} \frac{\sqrt{x+2} - 2}{\sqrt{x+7} - 3} = \lim_{x \to 2} \frac{\frac{x - 2}{\sqrt{x+2} + 2}}{\sqrt{x+7} - 3} \][/tex]
Next, handle the denominator [tex]\(\sqrt{x+7} - 3\)[/tex] by a similar rationalization:
[tex]\[ \sqrt{x+7} - 3 = \frac{(\sqrt{x+7} - 3)(\sqrt{x+7} + 3)}{\sqrt{x+7} + 3} = \frac{(x + 7) - 9}{\sqrt{x+7} + 3} = \frac{x - 2}{\sqrt{x+7} + 3} \][/tex]
Substitute this back into the modified expression:
[tex]\[ \frac{\frac{x - 2}{\sqrt{x+2} + 2}}{\frac{x - 2}{\sqrt{x+7} + 3}} = \lim_{x \to 2} \frac{x - 2}{\sqrt{x+2} + 2} \cdot \frac{\sqrt{x+7} + 3}{x - 2} \][/tex]
As the terms [tex]\(x - 2\)[/tex] in the numerator and denominator cancel out, we get:
[tex]\[ \lim_{x \to 2} \frac{\sqrt{x+7} + 3}{\sqrt{x+2} + 2} \][/tex]
Now we can substitute [tex]\(x = 2\)[/tex] directly into the simplified expression:
[tex]\[ \frac{\sqrt{2+7} + 3}{\sqrt{2+2} + 2} = \frac{\sqrt{9} + 3}{\sqrt{4} + 2} = \frac{3 + 3}{2 + 2} = \frac{6}{4} = \frac{3}{2} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 2}\left(\frac{\sqrt{x+2}-2}{\sqrt{x+7}-3}\right) = \frac{3}{2} \][/tex]
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