Problem 16 - Technical Mathematics with Calculus

Question

Limits Involving Zero or Infinity $$\lim _{x \rightarrow 0} \frac{3+x-x^{2}}{(x+3)(5-x)}$$

Step-by-Step Solution

Answer

\frac{1}{5}

Step 1: Understand the Limit

The first step is to understand the limit that we are trying to solve. We are asked to find the value of the function \( \frac{3+x-x^{2}}{(x+3)(5-x)} \) as \( x \) approaches 0. This involves evaluating the function at values close to 0 and seeing if it approaches a particular value.

Step 2: Substitute the value of x

Substitute \( x = 0 \) into the function to see if it directly gives us the limit without any indeterminate forms. \[ \lim _{x \rightarrow 0} \frac{3+x-x^{2}}{(x+3)(5-x)} = \frac{3+0-0^{2}}{(0+3)(5-0)} = \frac{3}{3 \times 5} = \frac{3}{15} = \frac{1}{5} \] Since substituting \( x = 0 \) into the function yields a finite value, there is no indeterminate form, and the limit can be directly calculated.

Step 3: Conclude the limit

As the direct substitution of \( x = 0 \) into the function gave us a finite value, there are no more steps required. The limit of the function as \( x \) approaches 0 is \( \frac{1}{5} \).

Key Concepts

Limit Evaluation

Limit evaluation is a fundamental concept in calculus that involves finding the value that a function approaches as the input gets close to some point. In our example, we are examining the behavior of the function \( \frac{3+x-x^2}{(x+3)(5-x)} \) as \( x \) approaches 0. This process aims to determine whether the function approaches a specific value, increases without bound, or fluctuates without settling on a single value as \( x \) gets arbitrarily close to 0.

To perform a limit evaluation, one generally follows three main steps. First, understand the problem presented and what is being asked. Second, apply the appropriate method to find the limit—if it exists. In our case, the direct substitution method was successful, revealing the limit as \( x \) approaches 0 to be \( \frac{1}{5} \). Lastly, confirm the result and ensure it makes sense in the context of the function's behavior around the point of interest.

Indeterminate Forms

Indeterminate forms occur when a limit presents an expression that is not initially clear or is undefined, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms make it impossible to determine the limit's value without further analysis. In our exercise, we encounter a fraction where both the numerator and the denominator tend to zero as \( x \) approaches 0, and it might seem at first that we are facing an indeterminate form. However, after applying the direct substitution method, we notice that both numerator and denominator give non-zero values, leading to a simple and well-defined result.

When faced with actual indeterminate forms, other techniques such as factoring, rationalizing, or using L'Hôpital's rule come into play. These methods help to simplify or transform the limit so that the indeterminate nature is resolved and an evaluation can be made.

Direct Substitution Method

The direct substitution method is one of the simplest and most intuitive approaches to evaluating limits. It entails substituting the value to which \( x \) is approaching directly into the function. If the result is a real number, then the limit is that number. When \( x \) was replaced with 0 in our exercise, the function yielded \( \frac{1}{5} \), denoting that the limit is indeed \( \frac{1}{5} \) as \( x \) approaches 0.

However, this method only works if the substitution leads to a defined value and not an indeterminate form. It is always the first method to try because of its simplicity, but one must be ready to apply more complex techniques if it fails. When successful, as in this case, it allows for a swift and efficient determination of the limit, saving time and effort.