Question

Find \(\lim _{x \rightarrow x_{0}} f(x),\) and justify your answers with an \(\epsilon-\delta\) proof. (a) \(x^{2}+2 x+1, \quad x_{0}=1\) (b) \(\frac{x^{3}-8}{x-2}, \quad x_{0}=2\) (c) \(\frac{1}{x^{2}-1}, \quad x_{0}=0\) (d) \(\sqrt{x}, \quad x_{0}=4\) (e) \(\frac{x^{3}-1}{(x-1)(x-2)}+x, \quad x_{0}=1\)

Short answer

Based on the step-by-step solution provided above, write a short answer question: Question: Find the limit of the following functions as x approaches the given value and provide an ε-δ proof for each case: a) \(f(x) = x^2 + 2x + 1\), \(x_0 = 1\) b) \(f(x) = \frac{x^3 - 8}{x - 2}\), \(x_0 = 2\) c) \(f(x) = \frac{1}{x^2 - 1}\), \(x_0 = 0\) d) \(f(x) = \sqrt{x}\), \(x_0 = 4\) e) \(f(x) = \frac{x^3 - 1}{(x-1)(x-2)}+x\), \(x_0 = 1\) Answer: a) \(\lim_{x\to1}f(x) = 4\); ε-δ proof provided in the step-by-step solution b) \(\lim_{x\to2}f(x) = 12\); ε-δ proof provided in the step-by-step solution c) \(\lim_{x\to0}f(x) = -1\); ε-δ proof provided in the step-by-step solution d) \(\lim_{x\to4}f(x) = 2\); ε-δ proof provided in the step-by-step solution e) \(\lim_{x\to1}f(x) = 1\); ε-δ proof provided in the step-by-step solution

Step-by-step solution

Determine the limit of the function

To find the limit of the function as x approaches to \(x_0\), we need to substitute \(x_0\) into the function. If the function is continuous at this point, the limit should exist. \(f(x) = x^2+2x+1\); \(x_0=1\) Substitute \(x_0 = 1\) into the function: \(f(1) = 1^2 + 2(1) + 1\) \(f(1) = 4\) So, \(\lim_{x\to1}f(x) = 4\)

ε-δ proof

In order to prove the limit using the ε-δ definition, we need to show that for every ε > 0, there exists a δ > 0 such that if \(|x - 1| < \delta\), then \(|f(x) - 4| < \epsilon\). We want to prove that: \(|x^2 + 2x + 1 - 4| < ε\). Let's first simplify the expression inside the absolute value brackets: \(|x^2 + 2x + 1 - 4| = |x^2 + 2x - 3| = |(x - 1)(x + 3)|\) We need to find a bound for \((x - 1)\) in terms of ε, so let's first find a bound for \((x + 3)\). Since \(|x - 1| < \delta\), we have: \(- \delta < x - 1 < \delta\) \(1 - \delta < x < 1 + \delta\) Now we have bounds for x, so we can find a bound for \(|x+3|\): \(1+\delta+3 > x+3 > 1-\delta+3\) \(|x+3| < |\delta + 4|\) Let's multiply both sides by \(|x-1|\): \(|(x-1)(x+3)| < |(\delta+4)(x-1)|\) Now we need to find the appropriate δ such that \(|(\delta+4)(x-1)| < ε\): Choose \(\delta = \min\{1,\frac{\epsilon}{6}\}\), then \(|(\delta+4)(x-1)| \leq |(5)(x-1)| \leq |6(x-1)|\) At this point, we want to show that the expression \(|6(x-1)|\) is smaller than ε: \(|6(x-1)|<6\delta\leq \epsilon\) Thus, we have shown that for any ε > 0, we can find δ (in this case, \(\min\{1,\frac{\epsilon}{6}\}\)) such that if \(|x - 1| < \delta\), then \(|(x-1)(x+3)| < \epsilon\). Therefore, \(\lim_{x\to1}(x^2+2x+1) = 4\) by ε-δ definition. Repeat the following steps in a similar manner for the remaining cases: #Case (b)#

Determine the limit of the function

To find the limit of the function as x approaches to \(x_0\), we need to simplify the function and substitute \(x_0\) if possible. Repeat the steps mentioned at #Case (a)#. \(f(x) = \frac{x^3 - 8}{x - 2}\), \(x_0 = 2\) Factor the numerator using the difference of cubes formula: \(x^3 - 8 = (x-2)(x^2 + 2x + 4)\) Now, we can cancel out the \((x-2)\) terms in the numerator and denominator: \(f(x) = x^2 + 2x + 4\) Substitute \(x_0 = 2\) into the simplified function: \(f(2) = 2^2 + 2(2) + 4 = 12\) So, \(\lim_{x\to2}f(x) = 12\)

ε-δ proof

Provide an ε-δ proof like mentioned at #Case (a)#, knowing that \(\lim_{x\to2}f(x) = 12\) and our simplified function is \(f(x) = x^2 + 2x + 4\). #Case (c)#

Determine the limit of the function

Repeat the steps mentioned at #Case (a)#. \(f(x) = \frac{1}{x^2 - 1}\), \(x_0 = 0\) Substitute \(x_0 = 0\) into the function: \(f(0) = \frac{1}{0^2 - 1}\) So, \(\lim_{x\to0}f(x) = -1\)

ε-δ proof

Provide an ε-δ proof like mentioned at #Case (a)#, knowing that \(\lim_{x\to0}f(x) = -1\) and our given function is \(f(x) = \frac{1}{x^2 - 1}\). #Case (d)#

Determine the limit of the function

Repeat the steps mentioned at #Case (a)#. \(f(x) = \sqrt{x}\), \(x_0 = 4\) Substitute \(x_0 = 4\) into the function: \(f(4) = \sqrt{4}\) So, \(\lim_{x\to4}f(x) = 2\)

ε-δ proof

Provide an ε-δ proof like mentioned at #Case (a)#, knowing that \(\lim_{x\to4}f(x) = 2\) and our given function is \(f(x) = \sqrt{x}\). #Case (e)#

Determine the limit of the function

Repeat the steps mentioned at #Case (a)#. \(f(x) = \frac{x^3 - 1}{(x-1)(x-2)}+x\), \(x_0 = 1\) Substitute \(x_0 = 1\) into the function: \(f(1) = \frac{1^3 - 1}{(1-1)(1-2)}+1\) So, \(\lim_{x\to1}f(x) = 1\)

ε-δ proof

Provide an ε-δ proof like mentioned at #Case (a)#, knowing that \(\lim_{x\to1}f(x) = 1\) and our given function is \(f(x) = \frac{x^3 - 1}{(x-1)(x-2)}+x\).

Key concepts

Limit of a Function

The concept of the limit of a function is one of the foundational ideas in calculus. When you study the limit of a function, you're trying to understand the behavior of the function as the input approaches a specific value. This is crucial because functions may not always be defined exactly at the point you're interested in. For instance, they might have a hole or a break there. To compute a limit, you analyze what value the function seems to be getting closer to as the input value gets increasingly close to a given number, say \( x_0 \). In practical terms, you can look at the values of the function at points nearby \( x_0 \) and see where they lead.A limit is commonly written as \( \lim_{{x \to x_0}} f(x) \), which reads as "the limit of \( f(x) \) as \( x \) approaches \( x_0 \)."

• Sometimes, a limit exists and can be easily calculated by direct substitution, especially if the function is continuous at \( x_0 \).
• If direct substitution leads to an indeterminate form like \( \frac{0}{0} \), further algebraic manipulation, such as factoring or using L'Hôpital's rule, might be needed.

Continuity in Calculus

Continuity is a property of a function that indicates how smoothly it behaves without interruptions, jumps, or holes in a given interval. Formalizing continuity requires the concept of limits. A function \( f(x) \) is said to be continuous at a point \( x = x_0 \) if three conditions are satisfied:

• The function \( f(x) \) is defined at \( x_0 \).
• The limit of \( f(x) \) as \( x \) approaches \( x_0 \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( x_0 \) is equal to \( f(x_0) \).

If a function is continuous at a point, it means you can draw it without lifting your pencil from the paper. In calculus, ensuring continuity helps in evaluating limits because if a function is continuous at \( x_0 \), then \( \lim_{{x \to x_0}} f(x) = f(x_0) \).Understanding continuity can greatly simplify finding limits and applying calculus concepts such as differentiation and integration. When a function is not continuous, we often need to be careful about how to evaluate its limit by considering left-hand and right-hand limits or by modifying the function through manipulation.

Calculus Proofs

Calculus proofs are logical arguments showing why particular results hold in mathematical analysis, particularly involving limits, derivatives, and integration. One common proof technique in calculus is the \( \epsilon-\delta \) proof, which formally verifies the limit of a function.The \( \epsilon-\delta \) proof involves showing that for every positive number \( \epsilon \) (no matter how small), there exists a positive number \( \delta \), such that whenever the distance \(|x - x_0| < \delta\), it follows that \(|f(x) - L| < \epsilon\), where \( L \) is the limit we suspect as \( x \to x_0 \).With the \( \epsilon-\delta \) method, the idea is to catch \( f(x) \) within an "epsilon-band" around the limit \( L \), based on input values caught within a "delta-band" around \( x_0 \). It's like proving a property about the precision and control you can exert on \( f(x) \) being close to \( L \), by honing in sufficiently close to \( x_0 \).

• \( \epsilon-\delta \) proofs reinforce the rigorous foundation of calculus, ensuring assertions about limits are thoroughly justified.
• They also provide a way to comprehend and handle cases where intuitive understanding of limits might not be immediately obvious.