To solve the limit [tex]\(\lim _{x \rightarrow 2} \frac{\sqrt{x+2}-2}{x-2}\)[/tex], let's go through it step-by-step.
1. Substitution: First, try direct substitution of [tex]\( x = 2 \)[/tex] into the function:
[tex]\[
\frac{\sqrt{2+2}-2}{2-2} = \frac{\sqrt{4}-2}{0} = \frac{2-2}{0} = \frac{0}{0}
\][/tex]
This gives us an indeterminate form [tex]\( \frac{0}{0} \)[/tex], so we need to find another way to evaluate the limit.
2. Rationalizing the numerator: To handle the indeterminate form, we use the technique of multiplying by the conjugate to simplify the expression. The conjugate of [tex]\(\sqrt{x+2} - 2\)[/tex] is [tex]\(\sqrt{x+2} + 2\)[/tex].
Multiply the numerator and the denominator by [tex]\(\sqrt{x+2} + 2\)[/tex]:
[tex]\[
\frac{\sqrt{x+2}-2}{x-2} \cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}
\][/tex]
3. Simplify: This simplifies the numerator as follows:
[tex]\[
\frac{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}{(x-2)(\sqrt{x+2}+2)} = \frac{(x+2) - 4}{(x-2)(\sqrt{x+2}+2)}
\][/tex]
[tex]\[
= \frac{x - 2}{(x-2)(\sqrt{x+2}+2)}
\][/tex]
Now we can cancel out the [tex]\((x-2)\)[/tex] term in the numerator and the denominator:
[tex]\[
= \frac{1}{\sqrt{x+2} + 2}
\][/tex]
4. Take the limit: Now we can take the limit as [tex]\( x \)[/tex] approaches 2:
[tex]\[
\lim_{x \to 2} \frac{1}{\sqrt{x+2} + 2}
\][/tex]
Substitute [tex]\( x = 2 \)[/tex] into the simplified expression:
[tex]\[
= \frac{1}{\sqrt{2+2} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{\frac{1}{4}}
\][/tex]
