We are asked to evaluate the following limit:

$\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 2} - \sqrt{3}}.$

Step 1: Substitution

First, try direct substitution $x = 1$:

$\frac{1 - 1}{\sqrt{1^2 + 2} - \sqrt{3}} = \frac{0}{\sqrt{3} - \sqrt{3}} = \frac{0}{0},$

This gives an indeterminate form $\frac{0}{0}$, so we need to apply algebraic manipulation to resolve this.

Step 2: Rationalize the denominator

We can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator:

$\frac{x - 1}{\sqrt{x^2 + 2} - \sqrt{3}} \times \frac{\sqrt{x^2 + 2} + \sqrt{3}}{\sqrt{x^2 + 2} + \sqrt{3}} = \frac{(x - 1)(\sqrt{x^2 + 2} + \sqrt{3})}{(x^2 + 2) - 3}.$

The denominator simplifies as follows:

$(x^2 + 2) - 3 = x^2 - 1 = (x - 1)(x + 1).$

So the expression becomes:

$\frac{(x - 1)(\sqrt{x^2 + 2} + \sqrt{3})}{(x - 1)(x + 1)}.$

Step 3: Simplify

We can cancel the factor $(x - 1)$ in the numerator and denominator (assuming $x \neq 1$):

$\frac{\sqrt{x^2 + 2} + \sqrt{3}}{x + 1}.$

Step 4: Substitution again

Now substitute $x = 1$ into the simplified expression:

$\frac{\sqrt{1^2 + 2} + \sqrt{3}}{1 + 1} = \frac{\sqrt{3} + \sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}.$

Thus, the value of the limit is $\sqrt{3}$.

Answer:

The correct answer is option E.

Would you like further details or have any questions? Here are five related questions to deepen understanding:

1. What is the importance of rationalizing in limits?
2. What are indeterminate forms, and how do they affect limit calculations?
3. Can L'Hopital's Rule be applied to this problem? Why or why not?
4. How does the cancellation of terms in limits work algebraically?
5. What are the common strategies for solving limits involving square roots?

Tip: When encountering indeterminate forms in limits, try algebraic manipulation like factorization or rationalization to simplify the expression before re-evaluating.