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Calculate the limit:

[tex]\[ \lim_{x \rightarrow 1} \frac{x-1}{\sqrt{x^2+3}-2} \][/tex]


Sagot :

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]
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