Sure, let's solve the limit [tex]\( \lim_{x \to 0} \frac{\sqrt{x + 3} - \sqrt{3}}{x} \)[/tex] step-by-step.
1. Identify the Indeterminate Form:
First, let's substitute [tex]\( x = 0 \)[/tex] directly into the expression to see what we get:
[tex]\[
\frac{\sqrt{0 + 3} - \sqrt{3}}{0} = \frac{\sqrt{3} - \sqrt{3}}{0} = \frac{0}{0}
\][/tex]
We obtain an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
2. Apply Algebraic Manipulation (Rationalization):
To resolve this indeterminate form, we can apply a trick called rationalization. Multiply the numerator and the denominator by the conjugate of the numerator:
[tex]\[
\frac{\sqrt{x + 3} - \sqrt{3}}{x} \cdot \frac{\sqrt{x + 3} + \sqrt{3}}{\sqrt{x + 3} + \sqrt{3}}
\][/tex]
3. Simplify the Expression:
This will result in the following expression:
[tex]\[
\frac{(\sqrt{x + 3} - \sqrt{3})(\sqrt{x + 3} + \sqrt{3})}{x (\sqrt{x + 3} + \sqrt{3})}
\][/tex]
The numerator simplifies using the difference of squares:
[tex]\[
(\sqrt{x + 3})^2 - (\sqrt{3})^2 = (x + 3) - 3 = x
\][/tex]
So, we have:
[tex]\[
\frac{x}{x (\sqrt{x + 3} + \sqrt{3})} = \frac{1}{\sqrt{x + 3} + \sqrt{3}}
\][/tex]
4. Evaluate the Limit:
Now we evaluate the limit as [tex]\( x \)[/tex] approaches 0:
[tex]\[
\lim_{x \to 0} \frac{1}{\sqrt{x + 3} + \sqrt{3}}
\][/tex]
Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[
\frac{1}{\sqrt{0 + 3} + \sqrt{3}} = \frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2\sqrt{3}}
\][/tex]
5. Simplify the Final Result:
Simplify the reciprocal expression:
[tex]\[
\frac{1}{2\sqrt{3}} = \frac{1}{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2} \cdot \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{6}
\][/tex]
Hence, the limit is:
[tex]\[
\lim_{x \to 0} \frac{\sqrt{x + 3} - \sqrt{3}}{x} = \frac{\sqrt{3}}{6}
\][/tex]
