Problem 17 - Calculus

Question

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 5} \frac{x^{2}+25}{x-5} $$

Step-by-Step Solution

Answer

The limit does not exist.

Step 1: Analyze the Expression

Identify the expression inside the limit: \(\frac{x^{2}+25}{x-5}\). As \(x\to 5\), the denominator approaches zero, making the expression undefined at \(x=5\). This suggests a possible discontinuity or indeterminate form.

Step 2: Simplify the Numerator

Rewrite the expression with its components. The numerator is \(x^2 + 25\), which cannot be factored to cancel out the \(x-5\) in the denominator directly. This distributively means no cancellation occurs to remove the indeterminate form as \(x \rightarrow 5\).

Step 3: Substitute the Limit Value Directly

Substitute \(x = 5\) directly into the simplified expression to check: \(\frac{5^2 + 25}{5-5} = \frac{50}{0}\). This forms a \(\frac{50}{0}\) scenario, indicating the limit does not exist.

Key Concepts

Understanding Indeterminate Forms

When we talk about limits in calculus, particularly when dealing with expressions like \(\frac{x^2 + 25}{x-5}\), we often encounter the term "indeterminate form." Indeterminate forms arise in limits when initial substitution leads to expressions that are mathematically undefined, like \(\frac{0}{0}\) or \(\frac{50}{0}\). These forms require special attention because they suggest the limit either does not exist or requires algebraic manipulation to evaluate.

• An expression like \(\frac{0}{0}\) suggests further simplification may reveal the actual limit, via factoring, multiplying by conjugates, or other techniques.
• However, when you have a division by zero \(\frac{a}{0}\) where \(a eq 0\), it leads to an undefined expression.

In our exercise, the form \(\frac{50}{0}\) appeared, indicating that the function does not have a meaningful limit at \(x=5\), as division by zero is undefined.

Exploring Discontinuity

Discontinuity in functions occurs when there are abrupt changes in function values, points where the function is not defined, or points where the limit does not match the function value. In the limit problem \(\lim_{x \rightarrow 5} \frac{x^2 + 25}{x-5}\), the function presents a discontinuity at \(x=5\).

• This discontinuity is a "removable" one if algebraic manipulation can annul the problematic denominator.
• Otherwise, it's a "non-removable" discontinuity, which means the function cannot be simplified enough to define a limit.

Our step-by-step did not find a viable specification at \(x=5\), leading to non-existence of the limit due to the problematic division \(\frac{50}{0}\) and thereby cementing the discontinuity.

Limit Substitution Explained

Limit substitution involves directly substituting the value \(x\) approaches into the function to find the limit. If substitution results in an expression like \(\frac{0}{0}\), further steps are necessary to simplify or manipulate the expression. If it yields a number for an undefined form like \(\frac{a}{0}\), it often shows us that the limit doesn’t exist.When solving \(\lim_{x \rightarrow 5} \frac{x^2 + 25}{x-5}\), directly plugging in 5 results in \(\frac{50}{0}\), an undefined expression. Hence, limit substitution failed, confirming that the function does not possess a limit at this point. Exploring alternative simplification methods may sometimes resolve this, but here, the original expression lacked such opportunities.