Problem 40 - Calculus of a Single Variable

Question

Using the \(\varepsilon-\delta\) Definition of Limit In Exercises \(37-48\) , find the limit \(L\) . Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L .\) $$ \lim _{x \rightarrow 3}\left(\frac{3}{4} x+1\right) $$

Step-by-Step Solution

Answer

The limit \(L\) of the function \(\frac{3}{4} x + 1\) as \(x\) approaches 3 is \(3.25\), and this is proven using the \(\varepsilon - \delta\) definition of limit by choosing \(\delta = \varepsilon / 0.75\).

Step 1: Finding the limit \(L\)

Finding the limit of a function as \(x\) approaches a certain value essentially means substituting that value into the function. In this case, substituting \(x = 3\) into \(\frac{3}{4} x + 1\), we get \(\frac{3}{4} \cdot 3 + 1 = 3.25\). Therefore, the limit \(L\) as \(x\) approaches 3 is \(L = 3.25\).

Step 2: Proving the limit using the \(\varepsilon - \delta\) definition

The \(\varepsilon - \delta\) definition of a limit states that for any number \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\), if \(0 < |x - 3| < \delta\), then \(|\frac{3}{4} x + 1 - L| < \varepsilon\). In simpler terms, this means that we can make the function as close as we want to \(L\) by taking \(x\) sufficiently close to 3. Let's choose \(\delta = \varepsilon / 0.75\). When \(0 < |x - 3| < \delta\), we have \(|\frac{3}{4}x + 1 - 3.25| = |0.75x - 2.25| = 0.75 |x - 3| < 0.75 \delta = \varepsilon\). Hence, the limit is proven to be \(L = 3.25\).

Key Concepts

Limit of a function

In calculus, the concept of the limit of a function helps us understand the behavior of functions as they approach a specific point. For our given exercise, we're interested in what happens to the function \( f(x) = \frac{3}{4}x + 1 \) as \( x \) gets closer and closer to 3. By substituting \( x = 3 \) into the function, we find that \( \frac{3}{4} \times 3 + 1 = 3.25 \). This simple substitution shows us that as \( x \) approaches 3, the value of the function approaches 3.25. The limit is essentially the value the function tends towards as we hone in on 3 from both sides, left and right.

Proving limits

The rigorous approach to proving the limit of a function relies on the \( \varepsilon-\delta \) definition. This definition is all about precision. It states that for every small positive number \( \varepsilon \) (no matter how tiny), there exists another small positive number \( \delta \). This \( \delta \) is crucial because it tells us how close \( x \) must get to the limit point, here 3, so that the output of the function is within \( \varepsilon \) distance of the limit, \( L = 3.25 \).To prove the limit is \( 3.25 \) using this definition, choose \( \delta = \varepsilon / 0.75 \). We then confirm that whenever the difference \( |x - 3| \) is smaller than \( \delta \), the difference \( |\frac{3}{4}x + 1 - 3.25| \) remains smaller than \( \varepsilon \). This careful control ensures that our function can be made arbitrarily close to 3.25 by simply bringing \( x \) close enough to 3.

Calculus concepts

Calculus gives us powerful tools to analyze how things change. The idea of limits is foundational in calculus. It deals with approaching values and allows us to define derivatives and integrals.

• Limits: Establish the concept of approaching a value.
• Derivatives: Use limits to define instantaneous rates of change.
• Integrals: Essentially sums that make use of limit concepts over an interval.

The \( \varepsilon-\delta \) method is a fundamental technique that demonstrates the precision of calculus. It shows how a function's behavior near a point leads us to understand continuity and differentiability.In our example, the limit concept combined with the \( \varepsilon-\delta \) definition provides a clear and precise way to express functions' behaviors near specific points, illustrating how calculus offers comprehensive insight into mathematical and real-world phenomena.