/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Gives a function \(f(x),\) a poi... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 35

Gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) $$0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon$$ $$f(x)=\sqrt{1-5 x}, \quad c=-3, \quad \epsilon=0.5$$

Short Answer

Expert verified
The limit \(L\) is 4, and \(\delta = 0.75\) satisfies the inequality.

Step by step solution

01

Determine the Limit L

We must first find the limit \(L\) of the function \(f(x) = \sqrt{1 - 5x}\) as \(x\) approaches \(c = -3\). To do so, simply substitute \(c\) into the function: \[ L = \lim_{x \to -3} \sqrt{1 - 5x} = \sqrt{1 - 5(-3)} = \sqrt{1 + 15} = \sqrt{16} = 4. \] Thus, \(L = 4\).
02

Set Up the Inequality for 未

We want to find \(\delta > 0\) such that if \(0 < |x + 3| < \delta\), then \(|f(x) - L| < \epsilon\). Using \(L = 4\) and \(\epsilon = 0.5\), we have: \[ |\sqrt{1 - 5x} - 4| < 0.5. \]
03

Solve the Inequality

Now, solve the inequality \(|\sqrt{1 - 5x} - 4| < 0.5\). This can be split into two inequalities: \[ -0.5 < \sqrt{1 - 5x} - 4 < 0.5. \] Thus, solving it we get: \[ 3.5 < \sqrt{1 - 5x} < 4.5. \]
04

Square Both Sides

By squaring all parts of the inequality, we can remove the square root: \[ 3.5^2 < 1 - 5x < 4.5^2. \] This results in \[ 12.25 < 1 - 5x < 20.25. \]
05

Solve for x

We now solve for \(x\): \[ 12.25 < 1 - 5x \quad \Rightarrow \quad 11.25 < -5x \quad \Rightarrow \quad x < -2.25, \] \[ 1 - 5x < 20.25 \quad \Rightarrow \quad -5x < 19.25 \quad \Rightarrow \quad x > -3.85. \]
06

Determine 未 Interval

The interval for \(x\) is \(-3.85 < x < -2.25\). So, the distance from \(-3\) to the endpoints defines the values of \(\delta\):\[ \delta = \min(3 - 2.25, 3.85 - 3) = \min(0.75, 0.85) = 0.75. \] Thus, \(\delta = 0.75\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Epsilon-Delta Definition
The epsilon-delta definition is a mathematical formalism used to prove the limit of a function. It establishes how close function values need to be to a particular limit, when the input values are near a certain point. Here's how it works:
  • Given a function, a point, and an epsilon (\(\epsilon\)), which represents the maximum allowed difference from the limit.
  • You're tasked to find a delta (\(\delta\)), highlighting the tolerance for how close the variable is to the point.
  • The core statement is: Given any \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\) satisfying \(0 < |x - c| < \delta\), it holds that \(|f(x) - L| < \epsilon\).
This definition rigorously captures the intuitive idea that the function values \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(c\). In our particular exercise, after calculating, the appropriate \(\delta\) was found to be 0.75, ensuring \(|\sqrt{1 - 5x} - 4| < 0.5\).
Continuity
Continuity is a cornerstone concept in calculus, describing a function that is smooth without any breaks or jumps. If a function is continuous at a point, we can rely on its graph being unbroken at that point. Here's what defines continuity:
  • A function \(f(x)\) is continuous at a point \(c\) if the limit of \(f(x)\) as \(x\) approaches \(c\) from either side equals the function value \(f(c)\).
  • In mathematical form: \(\lim_{x \to c} f(x) = f(c)\).
For our function \(f(x) = \sqrt{1 - 5x}\), since we found a proper epsilon-delta relationship, it suggests continuity for the points considered. Additionally, because the square root function is continuous wherever it's defined (when the expression under the square root is positive), it reassures us about the function's behavior around the point \(c = -3\).
Square Roots
The square root function, denoted \(\sqrt{x}\), is very common in mathematics, involving finding a number that, when multiplied by itself, equals \(x\). The operation has some specific characteristics:
  • The square root is only defined for non-negative numbers in the real number system, as no real number squared will give a negative result.
  • It's typically an increasing function, meaning as the input increases, the output does as well.
  • In the exercise, we considered \(\sqrt{1 - 5x}\); the expression inside determines real values of \(x\) for which the function is valid.
In practical terms, the square root transformation commonly appears in equations that model real-world phenomena, such as physics or engineering contexts. For our specific example, understanding the range in which the function \(\sqrt{1 - 5x}\) remains valid guides the solution for identifying the appropriate \(\delta\) as \(0.75\). This ensures that \(|x + 3|<\delta\) holds, keeping \(|\sqrt{1 - 5x} - 4| < 0.5\).

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Most popular questions from this chapter

You will find a graphing calculator useful for Exercise. Let \(f(x)=\left(x^{2}-1\right) /(|x|-1).\) a. Make tables of the values of \(f\) at values of \(x\) that approach \(c=-1\) from above and below. Then estimate \(\lim _{x \rightarrow-1} f(x).\) b. Support your conclusion in part (a) by graphing \(f\) near \(c=-1\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow-1.\) c. Find \(\lim _{x \rightarrow-1} f(x)\) algebraically.

You will find a graphing calculator useful for Exercise. Let \(G(t)=(1-\cos t) / t^{2}.\) a. Make tables of values of \(G\) at values of \(t\) that approach \(t_{0}=0\) from above and below. Then estimate \(\lim _{t \rightarrow 0} G(t).\) b. Support your conclusion in part (a) by graphing \(G\) near \(t_{0}=0.\)

Explain why the equation \(\cos x=x\) has at least one solution.

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. Evaluate this limit for the given value of \(x\) and function \(f.\) $$f(x)=1 / x, \quad x=-2$$

Suppose that \(\lim _{x \rightarrow-2} p(x)=4, \lim _{x \rightarrow-2} r(x)=0,\) and \(\lim _{x \rightarrow-2} s(x)=-3 .\) Find a. \(\lim _{x \rightarrow-2}(p(x)+r(x)+s(x))\) b. \(\lim _{x \rightarrow-2} p(x) \cdot r(x) \cdot s(x)\) c. \(\lim _{x \rightarrow-2}(-4 p(x)+5 r(x)) / s(x)\)
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