Problem 21 - The Calculus with Analytic Geometry

Question

Prove that $$ \lim _{x \rightarrow-\infty} \frac{8 x+3}{2 x-1}=4 $$ by showing that for any \(\epsilon>0\) there exists a number \(N<0\) such that $$ \left|\frac{8 x+3}{2 x-1}-4\right|<\epsilon $$ whenever \(x

Step-by-Step Solution

Answer

\(N = \frac{-\frac{7}{\epsilon}+1}{2}\)

Step 1 - Simplify the expression

Start by rewriting the expression inside the limit. Subtract 4 from the given fraction and simplify.\(\left| \frac{8x+3}{2x-1} - 4 \right| = \left| \frac{8x+3 - 4(2x-1)}{2x-1} \right|\)

Step 2 - Expand and combine like terms

Expand and simplify the numerator:\(8x + 3 - 4(2x - 1) = 8x + 3 - 8x + 4 = 7\).The expression becomes:\(\left| \frac{7}{2x-1} \right|\).

Step 3 - Establish the condition for \epsilon

We need to find a negative number \(N\) such that:\(\left| \frac{7}{2x-1} \right| < \epsilon\) for all \(x < N\).

Step 4 - Solve for \N

To ensure the condition holds, solve the inequality:\(\left| \frac{7}{2x-1} \right| < \epsilon\).This simplifies to:\(\frac{7}{|2x-1|} < \epsilon\), which is equivalent to:\(|2x-1| > \frac{7}{\epsilon}\).

Step 5 - Determine the critical value of \x

Since \(x\) is approaching negative infinity, solve for \(x\):\(2x-1 < -\frac{7}{\epsilon}\)\(2x < -\frac{7}{\epsilon}+1\)\(x < \frac{-\frac{7}{\epsilon}+1}{2}\).This gives the value of \(N = \frac{-\frac{7}{\epsilon}+1}{2}\).

Step 6 - Conclude the proof

Thus, for any \(\epsilon > 0\), there exists a number \(N = \frac{-\frac{7}{\epsilon}+1}{2}\) such that whenever \(x < N\), the inequality \(\left| \frac{8x+3}{2x-1} - 4 \right| < \epsilon\) will hold true.

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to define the limit of a function. It provides a rigorous way to say that a function approaches a certain value as the input approaches some point. In this context, our task is to prove that \(\lim_{x \rightarrow -\infty} \frac{8x+3}{2x-1}=4\).
To start, we need to demonstrate that for any \( \epsilon > 0\), there is a number \( N < 0\)} such that whenever \( x < N \) It helps us show the limit by bounding the function’s distance from the limit by a small positive number \( \epsilon\). This means we want:\(\left|\frac{8x+3}{2x-1}-4\right|<\epsilon\).
By breaking down the function and manipulating it algebraically, we can specify the conditions under which this inequality holds. This concept forms the foundation of proving limits rigorously.

Asymptotic Behavior

Asymptotic behavior refers to how a function behaves as the input approaches large values, such as positive or negative infinity. In this problem, we consider \(\lim_{x \rightarrow -\infty}\) because we are interested in the behavior of \( \frac{8x+3}{2x-1}\) as \( x \) becomes negatively large.
This tells us that beyond a certain (very negative) value of \( x\), the value of \(\frac{8x+3}{2x-1}\) gets closer and closer to \( 4 \), without ever touching it.
Think of it as the curve getting very close to a horizontal line (the asymptote), helping us understand the long-term behavior of the function. Before simplifying the fractions, the proof was only a symbolic representation, but after simplification, we can visualize how the function approaches 4.

Algebraic Manipulation

Algebraic manipulation allows us to simplify expressions and solve equations systematically. This particular problem relies heavily on it, especially in steps like:\(\frac{8x+3}{2x-1} - 4\).
Substituting \(4 \) with \( \frac{8x-4}{2x -1}\) helps simplify to:\(\frac{8x + 3 - 4(2x -1)}{2x -1}\).
Breaking it down further leads to:\( \left| \frac{7}{2 x - 1} \right|\),which makes it easier to solve for the critical value of \( x \) that satisfies the inequality. Algebraic steps make the solution clearer and more straightforward, handling every component and reassembling them to prove the desired limit.