Question

The limit at infinity \(\lim _{x \rightarrow \infty} f(x)=L\) means that for any \(\varepsilon>0\) there exists \(N>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad x>N.$$ Use this definition to prove the following statements. $$\lim _{x \rightarrow \infty} \frac{2 x+1}{x}=2$$

Short answer

Question: Prove that \(\lim_{x \rightarrow \infty} \frac{2x + 1}{x} = 2\) using the definition of the limit at infinity. Answer: We found that for any \(\varepsilon > 0\), we can choose \(N = \frac{1}{\varepsilon}\), such that for all \(x > N\), we have \(\left|\frac{2x + 1}{x} - 2\right| < \varepsilon\). Hence, we have proven that \(\lim_{x \rightarrow \infty} \frac{2x + 1}{x} = 2\).

Step-by-step solution

Define the function and the given limit

We are given the function $$f(x) = \frac{2x + 1}{x}$$ and the limit \(L = 2\).

Start by stating the condition given by the definition

According to the definition, for any \(\varepsilon > 0\), we need to find \(N > 0\) such that $$|f(x) - L| < \varepsilon$$ for all \(x > N\).

Substitute the function f(x) and the limit L into the inequality

We know the function \(f(x)\) and the limit \(L\), so we can substitute them into the inequality. $$\left|\frac{2x + 1}{x} - 2\right| < \varepsilon$$

Simplify the expression inside the absolute value

Let's simplify the expression inside the absolute value. $$\left|\frac{2x + 1 - 2x}{x}\right| = \left|\frac{1}{x}\right| < \varepsilon$$

Find a suitable value for N

Since \(x > 0\), we have \(\frac{1}{x} > 0\). We can rewrite the inequality as $$\frac{1}{x} < \varepsilon$$ Multiplying both sides by \(x\) and dividing both sides by \(\varepsilon\) gives $$x > \frac{1}{\varepsilon}$$ We can choose \(N = \frac{1}{\varepsilon}\) as the suitable value for N.

Conclusion

For any \(\varepsilon > 0\), we find \(N = \frac{1}{\varepsilon}\) such that for all \(x > N\), we have $$\left|\frac{2x + 1}{x} - 2\right| < \varepsilon$$ Therefore, we have proven that $$\lim _{x \rightarrow \infty} \frac{2 x+1}{x}=2$$

Key concepts

Epsilon-Delta Definition of Limit

Understanding the \textbf{epsilon-delta definition of limit} is crucial to studying calculus. At its core, this definition gives us a formal way of grasping what it means for a function to approach a specific value, known as the limit, as the input (or the variable) approaches a certain point.

In the context of a limit at infinity, the definition adapts to describe how a function behaves as the input grows without bound. We denote this as \( \textstyle\text{if} \lim_{x \rightarrow \infty} f(x) = L \) then for every \(\epsilon > 0\), there's a number \(N\) such that \( |f(x) - L| < \varepsilon \) whenever \( x > N \). In simpler terms, no matter how small a positive number (\(\epsilon\)) you choose, the function's value \(f(x)\) will be within that \(\epsilon\) distance from \(L\) once \(x\) is sufficiently large.

This forms the backbone for proving the behavior of functions at infinity, as it provides a mathematical way to show that a function gets arbitrarily close to a limit.

Limit of a Rational Function

The \textbf{limit of a rational function} as \(x\) approaches infinity looks at the end behavior of a fraction where both the numerator and denominator are polynomials. In our example, the function \(f(x) = \frac{2x + 1}{x}\) is rational. As \(x\) becomes very large, the \(1\) in the numerator becomes negligible compared to the terms with \(x\), so the function behaves like \(\frac{2x}{x}\).

This simplifies to \(2\), giving us an intuitive guess that the limit might be \(2\) as \(x\) approaches infinity. To prove this rigorously, we apply the epsilon-delta definition. By following a series of algebraic steps to express the function's deviation from \(2\) and setting it to be less than an arbitrary \(\varepsilon\), we can solve for \(N\) that validates our prediction.

Asymptotic Behavior

The term \textbf{asymptotic behavior} refers to how a function behaves as it moves towards infinity or some boundary value. In many cases, this is connected to finding the end behavior of a function or understanding how it stabilizes or fluctuates as its input grows large.

Our exercise focuses on a very common asymptotic behavior: stabilization towards a horizontal asymptote. For \(f(x) = \frac{2x + 1}{x}\), as \(x\) increases, the term \(\frac{1}{x}\) approaches \(0\) and the function itself approaches \(2\), making \(y = 2\) a horizontal asymptote. This is reflected graphically and can be proven with the epsilon-delta definition, underscoring the function's predictable behavior as \(x\) becomes very large.

Proof in Calculus

In the realm of calculus, a \textbf{proof} is a logical argument that demonstrates the truth of a mathematical statement. Proofs employ definitions, theorems, and established results to justify each step leading to the conclusion.

In the case of proving a limit, the proof must show that no matter how close you want the function's value to get to the predicted limit (when the variable goes towards infinity), there is always a point after which all the function's values will stay within the chosen closeness (denoted by \(\epsilon\)). The rigorous process for doing this often involves manipulating inequalities, as seen in the step-by-step solution we discussed. This form of proof is essential, as it provides a solid foundation to the kind of rigorous argumentation required in higher mathematics.