Problem 62 - Precalculus

Question

Evaluate the limits using the limit properties. $$\lim _{x \rightarrow 2} \frac{11-3 x^{2}}{\sqrt{x^{2}+3 x-1}}$$

Step-by-Step Solution

Answer

The limit is \( \frac{-1}{3} \).

Step 1: Direct Substitution

To evaluate the limit \( \lim _{x \rightarrow 2} \frac{11-3 x^{2}}{\sqrt{x^{2}+3 x-1}} \), first try to substitute \( x = 2 \) directly into the expression: \[\frac{11 - 3 \times 2^2}{\sqrt{2^2 + 3 \times 2 - 1}}\]Upon computation, this gives:\[\frac{11 - 12}{\sqrt{4 + 6 - 1}} = \frac{-1}{\sqrt{9}} = \frac{-1}{3}\]

Step 2: Verify Continuity

Check for any discontinuity at \( x = 2 \) by ensuring the expression doesn't lead to an indeterminate form. The numerator \( 11 - 3x^2 \) and the denominator \( \sqrt{x^2 + 3x - 1} \) are both continuous at \( x = 2 \).As calculated, no division by zero or any other undefined operations occur, confirming continuity.

Step 3: Conclude the Limit

Since substituting \( x = 2 \) into the expression results in a finite number and not an indeterminate form, the limit is equal to the result from direct substitution. Therefore, the limit is \( \frac{-1}{3} \).

Key Concepts

Direct Substitution

When tackling limit problems, direct substitution is often the first method we try. It involves directly replacing the variable in the limit with the value it approaches. For instance, in the problem \( \lim _{x \rightarrow 2} \frac{11-3 x^{2}}{\sqrt{x^{2}+3 x-1}} \), we substitute \( x = 2 \) into the expression. This results in computing: \[\frac{11 - 3 \times 2^2}{\sqrt{2^2 + 3 \times 2 - 1}} = \frac{11 - 12}{\sqrt{4 + 6 - 1}} = \frac{-1}{\sqrt{9}} = \frac{-1}{3}\] Direct substitution works best when the expression is genuinely continuous around the point being evaluated and does not result in an indeterminate form. If successful, as it is in this case, it provides a quick and straightforward solution.

Continuity

Continuity at a point means that the function doesn't have any breaks, jumps, or holes at that particular point. For a limit problem, confirming the continuity of the function at the given point helps assure us that the direct substitution is valid.To confirm continuity in our example, we need to check both the numerator \(11 - 3x^2\) and the denominator \(\sqrt{x^2 + 3x - 1}\) at \(x = 2\). Both parts evaluate to finite, non-zero quantities, meaning they are continuous there. Importantly:

• The numerator, \(11 - 3 \times 2^2\), evaluates to \(-1\), which is finite.
• The denominator \(\sqrt{4 + 6 - 1}\) evaluates to \(3\), which is also a non-zero finite number.

This continuous behavior around \(x = 2\) ensures no strange behavior that might disrupt a simple direct substitution.

Indeterminate Forms

Indeterminate forms are expressions where the limit cannot be directly computed using substitution due to mathematical uncertainties. Commonly seen forms include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). These require more advanced techniques, such as L'Hôpital's rule or algebraic manipulations, to resolve.In the original problem, by substituting \(x = 2\), we found a clean result of \(\frac{-1}{3}\). This indicates that the expression does not belong to an indeterminate form category. Thus, the limit can be simply determined using substitution, avoiding the need for more complicated procedures.Understanding indeterminate forms is vital because they help identify when a limit problem might need more than just direct substitution to be solved efficiently and correctly. Knowing how to recognize and resolve these forms will enhance your problem-solving toolkit in calculus.