Problem 51 - Precalculus

Question

Evaluate the limits using the limit properties. $$\lim _{x \rightarrow 2} \frac{2 x^{2}-5}{x+3}$$

Step-by-Step Solution

Answer

The limit is \( \frac{3}{5} \).

Step 1: Apply Direct Substitution

To find the limit \( \lim _{x \rightarrow 2} \frac{2x^{2} - 5}{x + 3} \), first try to evaluate it by directly substituting \( x = 2 \) into the expression. This yields \( \frac{2(2)^2 - 5}{2 + 3} = \frac{2 \cdot 4 - 5}{5} = \frac{8 - 5}{5} = \frac{3}{5} \).

Step 2: Verify Continuity

The function \( f(x) = \frac{2x^{2} - 5}{x + 3} \) is a rational function. Rational functions are continuous everywhere in their domain. Since there's no zero denominator for \( x = 2 \), the function is continuous at this point.

Step 3: Conclude the Limit

Given that direct substitution resulted in a finite number and the function is continuous at \( x = 2 \), the limit is confirmed to be \( \frac{3}{5} \). Therefore, \( \lim _{x \rightarrow 2} \frac{2x^{2} - 5}{x + 3} = \frac{3}{5} \).

Key Concepts

Direct Substitution

When evaluating limits, direct substitution can be a straightforward approach. This technique involves simply plugging the value that the variable approaches directly into the function. For the problem at hand, we have the limit \( \lim _{x \rightarrow 2} \frac{2x^{2} - 5}{x + 3} \).
By substituting \( x = 2 \) into the expression, we perform the following calculation:

• \( 2(2)^2 = 8 \)
• The numerator then becomes \( 8 - 5 = 3 \)
• The denominator is \( 2 + 3 = 5 \)
• Thus, the expression evaluates to \( \frac{3}{5} \)

Direct substitution is only valid when the function is defined and the denominator does not approach zero. It is always good practice to verify any indeterminate form before applying this method.

Continuity

A function is considered continuous at a point if it doesn't have any abrupt jumps or breaks there. More precisely, a function \( f(x) \) is continuous at a point \( x = a \) if:

• \( f(a) \) is defined
• \( \lim_{x \to a} f(x) \) exists
• \( \lim_{x \to a} f(x) = f(a) \)

For rational functions like \( f(x) = \frac{2x^2 - 5}{x + 3} \), continuity can be expected anywhere in the function's domain, which excludes points where the denominator is zero. In this exercise, since \( x = 2 \) does not cause the denominator to become zero, the function is continuous at this point. This confirms that we can confidently apply direct substitution.

Rational Functions

Rational functions are expressed as the quotient of two polynomials, like our example \( \frac{2x^2 - 5}{x + 3} \). These functions have unique characteristics influencing their continuity and differentiability.
Some key points to note about rational functions are:

• They are continuous at all points in their domain.
• The domain includes all real numbers except where the denominator equals zero.
• If a rational function has no shared factors between the numerator and denominator, it's fully simplified.

For our specific function, the domain is all real numbers since \( x + 3 eq 0 \). Thus, when evaluating limits using rational functions, it is crucial to identify these domain restrictions to ensure accurate calculations. Only then can techniques like direct substitution be applied effectively.