To find the limit [tex]\(\lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x^2+3}-2}\)[/tex], we will proceed through a series of algebraic manipulations and limit properties. Here is a detailed step-by-step solution:
1. Substitute [tex]\(x = 1\)[/tex] to identify the indeterminate form:
[tex]\( \frac{1-1}{\sqrt{1^2+3} - 2} = \frac{0}{\sqrt{4} - 2} = \frac{0}{2 - 2} = \frac{0}{0} \)[/tex]
This is an indeterminate form, [tex]\( \frac{0}{0} \)[/tex]. So, we need to proceed further using algebraic methods such as rationalizing the denominator.
2. Rationalize the denominator:
To rationalize, we multiply the numerator and the denominator by the conjugate of the denominator, which is [tex]\( \sqrt{x^2 + 3} + 2 \)[/tex]:
[tex]\[
\frac{x - 1}{\sqrt{x^2 + 3} - 2} \cdot \frac{\sqrt{x^2 + 3} + 2}{\sqrt{x^2 + 3} + 2}
\][/tex]
This yields:
[tex]\[
\frac{(x - 1)(\sqrt{x^2 + 3} + 2)}{(\sqrt{x^2 + 3} - 2)(\sqrt{x^2 + 3} + 2)}
\][/tex]
3. Simplify the expression:
The denominator will now simplify as a difference of squares:
[tex]\[
(\sqrt{x^2 + 3} - 2)(\sqrt{x^2 + 3} + 2) = (\sqrt{x^2 + 3})^2 - 2^2 = x^2 + 3 - 4 = x^2 - 1
\][/tex]
Therefore, the expression becomes:
[tex]\[
\frac{(x - 1)(\sqrt{x^2 + 3} + 2)}{x^2 - 1}
\][/tex]
4. Factorize the denominator:
Notice that [tex]\( x^2 - 1 \)[/tex] can be factored as a difference of squares:
[tex]\[
x^2 - 1 = (x -1)(x + 1)
\][/tex]
Hence, the expression now becomes:
[tex]\[
\frac{(x - 1)(\sqrt{x^2 + 3} + 2)}{(x - 1)(x + 1)}
\][/tex]
For [tex]\(x \neq 1\)[/tex], [tex]\( (x - 1) \)[/tex] terms cancel each other out:
[tex]\[
\frac{\sqrt{x^2 + 3} + 2}{x + 1}
\][/tex]
5. Compute the limit:
Now, we can substitute [tex]\(x = 1\)[/tex] into the simplified expression:
[tex]\[
\lim _{ x \rightarrow 1} \frac{\sqrt{x^2 + 3} + 2}{x + 1} = \frac{\sqrt{1^2 + 3} + 2}{1 + 1} = \frac{\sqrt{4} + 2}{2} = \frac{2 + 2}{2} = \frac{4}{2} = 2
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{2}
\][/tex]
