To evaluate the limit

$\lim_{{x \to 0}} \frac{3 - \sqrt{x^2 + 9}}{x^2},$

we notice that directly substituting $x = 0$ leads to the indeterminate form $\frac{0}{0}$. To resolve this, we can use algebraic manipulation to simplify the expression.

Step-by-Step Solution

1. Rationalize the numerator: To simplify the expression, we multiply the numerator and the denominator by the conjugate of the numerator, $3 + \sqrt{x^2 + 9}$. This helps eliminate the square root in the numerator:

   $\lim_{{x \to 0}} \frac{(3 - \sqrt{x^2 + 9})(3 + \sqrt{x^2 + 9})}{x^2 (3 + \sqrt{x^2 + 9})}.$

2. Simplify the expression: Use the difference of squares formula in the numerator:

   $(3 - \sqrt{x^2 + 9})(3 + \sqrt{x^2 + 9}) = 3^2 - (\sqrt{x^2 + 9})^2 = 9 - (x^2 + 9).$

   This simplifies to:

   $9 - x^2 - 9 = -x^2.$

   The expression now becomes:

   $\lim_{{x \to 0}} \frac{-x^2}{x^2 (3 + \sqrt{x^2 + 9})}.$

3. Cancel out common terms: The $x^2$ in the numerator and denominator can be canceled (as long as $x \neq 0$):

   $\lim_{{x \to 0}} \frac{-1}{3 + \sqrt{x^2 + 9}}.$

4. Evaluate the limit: Now, we substitute $x = 0$ into the simplified expression:

   $\frac{-1}{3 + \sqrt{0^2 + 9}} = \frac{-1}{3 + 3} = \frac{-1}{6}.$

Therefore, the limit is

$\lim_{{x \to 0}} \frac{3 - \sqrt{x^2 + 9}}{x^2} = -\frac{1}{6}.$

Would you like more details on any part of this solution or have any questions?

Related Questions

1. How can conjugates be used to simplify limits involving square roots?
2. What are other methods to solve limits that result in indeterminate forms?
3. How do you evaluate $\lim_{{x \to 1}} \frac{1 - \sqrt{x}}{x - 1}$?
4. What role does factoring play in simplifying expressions for limit calculations?
5. How does L'Hopital's Rule apply to limits involving indeterminate forms?

Tip: When dealing with square roots in limits, multiplying by the conjugate can help simplify the expression and eliminate the indeterminate form.