Question

Finding a Limit In Exercises \(5-22,\) find the limit. $$ \lim _{x \rightarrow 3} \frac{\sqrt{x+6}}{x+2} $$

Short answer

The limit as x approaches 3 of the function \( \frac{\sqrt{x+6}}{x+2} \) is \( \frac{3}{5} \).

Step-by-step solution

Direct Substitution

First, attempt to substitute x = 3 directly into the function. Here is what that would look like: \[ \lim _{x \rightarrow 3} \frac{\sqrt{x+6}}{x+2} = \frac{\sqrt{3+6}}{3+2} \] This produces the result \( \frac{\sqrt{9}}{5} \), or \( \frac{3}{5} \). However, since this did not result in an undetermined form, we do not need to perform additional steps to find the limit.

Verifying the Result

As a final step, confirm the correctness of your answer. Try evaluating the function at points closer and closer to 3 (both smaller and larger than 3) and ensure that the function value gets closer and closer to \( \frac{3}{5} \).

Key concepts

Limit of a Function

Understanding the limit of a function is a foundational concept in calculus. A limit describes the value that a function approaches as the input (or 'x' value) gets closer to some number. Limits are essential for studying the behavior of functions as they approach specific points, even if the function itself is not defined at that exact point.

For example, consider the exercise \( \lim _{x \rightarrow 3} \frac{\sqrt{x+6}}{x+2} \). Here, we're interested in the behavior of the function as 'x' approaches 3, not necessarily the function's value at 'x' = 3. This reflects a key aspect of calculus, the predictive analysis of functions' behaviors around certain values. By investigating limits, we can uncover continuity, derivatives, and the overall behavior of functions without precise point evaluations.

Direct Substitution in Limits

Direct substitution is often the first method attempted when evaluating the limit of a function as 'x' approaches a particular value. This approach is straightforward: you simply plug the target value into the function, provided the function is well-defined at that point. If direct substitution yields a definite number (as opposed to an indeterminate form like 0/0 or ∞/∞), then that number is the limit.

In the given problem, we apply direct substitution to find \( \lim _{x \rightarrow 3} \frac{\sqrt{x+6}}{x+2} \) and discover that the function simplifies to \( \frac{\sqrt{9}}{5} \) or \( \frac{3}{5} \). This quick and efficient step reveals the function's limit without the need for complex calculations or adjustments to the function. It's a critical tool for any student studying calculus.

Calculus of a Single Variable

Calculus of a single variable is essentially the calculus that deals with functions of one variable. It includes foundational operations such as finding limits, derivatives, and integrals of these functions. The problems focus on how a variable changes in one dimension and how this affects the function's value.

When we work on finding the limit as in the given exercise, we're engaging with a critical part of single-variable calculus. Each limit computation informs us about the change in function behavior, which is integral to understanding continuity and eventual derivative concepts. It's a step-by-step approach to mastering how a function behaves as its inputs get infinitely close to a certain point, ultimately helping students understand deeper concepts in calculus.
In essence, studying the calculus of a single variable sets the groundwork for all future work in the discipline, providing the necessary tools to navigate the dynamic landscape of continuous functions.