To determine the limit [tex]\(\lim_{x \rightarrow 0} \frac{\sqrt{a+x} - \sqrt{a-x}}{x}\)[/tex], we can proceed through a series of steps to simplify the expression and find the limit. Here's a step-by-step solution:
1. Rationalize the Numerator:
To simplify the expression, we can first rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator. The conjugate of [tex]\(\sqrt{a+x} - \sqrt{a-x}\)[/tex] is [tex]\(\sqrt{a+x} + \sqrt{a-x}\)[/tex]. Thus, we have:
[tex]\[
\frac{\sqrt{a+x} - \sqrt{a-x}}{x} \cdot \frac{\sqrt{a+x} + \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}} = \frac{(\sqrt{a+x} - \sqrt{a-x})(\sqrt{a+x} + \sqrt{a-x})}{x (\sqrt{a+x} + \sqrt{a-x})}
\][/tex]
2. Simplify the Numerator:
The numerator simplifies as follows:
[tex]\[
(\sqrt{a+x} - \sqrt{a-x})(\sqrt{a+x} + \sqrt{a-x}) = (\sqrt{a+x})^2 - (\sqrt{a-x})^2 = (a+x) - (a-x) = a+x - a + x = 2x
\][/tex]
3. Simplify the Whole Expression:
Now, substituting [tex]\(2x\)[/tex] back into the expression, we get:
[tex]\[
\frac{2x}{x (\sqrt{a+x} + \sqrt{a-x})} = \frac{2x}{x} \cdot \frac{1}{\sqrt{a+x} + \sqrt{a-x}} = \frac{2}{\sqrt{a+x} + \sqrt{a-x}}
\][/tex]
4. Calculate the Limit:
We need to find the limit of this simplified expression as [tex]\(x \rightarrow 0\)[/tex]:
[tex]\[
\lim_{x \rightarrow 0} \frac{2}{\sqrt{a+x} + \sqrt{a-x}}
\][/tex]
As [tex]\(x\)[/tex] approaches 0, [tex]\(\sqrt{a+x}\)[/tex] and [tex]\(\sqrt{a-x}\)[/tex] both approach [tex]\(\sqrt{a}\)[/tex]. Therefore,
[tex]\[
\sqrt{a+x} + \sqrt{a-x} \rightarrow \sqrt{a} + \sqrt{a} = 2\sqrt{a}
\][/tex]
Thus, the limit of the expression is:
[tex]\[
\lim_{x \rightarrow 0} \frac{2}{\sqrt{a+x} + \sqrt{a-x}} = \frac{2}{2\sqrt{a}} = \frac{1}{\sqrt{a}}
\][/tex]
In conclusion, the limit is:
[tex]\[
\boxed{\frac{1}{\sqrt{a}}}
\][/tex]
