Problem 54 For the limit $$\lim _{x \righ... [FREE SOLUTION]

Chapter 1: Problem 54

For the limit $$\lim _{x \rightarrow \infty} \frac{\sqrt{4 x^{2}+1}}{x+1}=2$$ illustrate Definition 7 by finding values of $N$ that correspond to $\varepsilon=0.5$ and $\varepsilon=0.1$

Short Answer

Expert verified

For $\varepsilon = 0.5$, $N = 1$. For $\varepsilon = 0.1$, $N = 9$.

Step by step solution

Understand the Definition of a Limit

According to Definition 7, the limit $ \lim_{x \to \infty} f(x) = L $ implies that for every $ \varepsilon > 0 $, there exists a number $ N $ such that for all $ x > N $, $ |f(x) - L| < \varepsilon $. In this exercise, $ f(x) = \frac{\sqrt{4x^2+1}}{x+1} $ and $ L = 2 $. We need to find values of $ N $ for given $ \varepsilon $.

Simplify the Expression for f(x)

Simplify the expression to check the behavior as $ x \to \infty $. Start by considering:\[\frac{\sqrt{4x^2+1}}{x+1} = \frac{\sqrt{4x^2+1}}{x(1 + \frac{1}{x})} = \frac{\sqrt{4 + \frac{1}{x^2}}}{1 + \frac{1}{x}}\]As $ x \to \infty $, $ \frac{1}{x^2} \to 0 $ and $ \frac{1}{x} \to 0 $, so the expression simplifies to $ \frac{\sqrt{4}}{1} = 2 $.

Establish the Inequality |f(x) - L| < 蔚

We need to solve $ \left| \frac{\sqrt{4x^2+1}}{x+1} - 2 \right| < \varepsilon $. Simplify this:\[\left| \frac{\sqrt{4x^2+1}}{x+1} - 2 \right| = \left| \frac{\sqrt{4x^2+1} - 2(x+1)}{x+1} \right| = \left| \frac{\sqrt{4x^2 + 1} - 2x - 2}{x+1} \right|\]

Find Critical Value of N for 蔚 = 0.5

Find $ N $ such that $ \left| \frac{\sqrt{4x^2+1} - 2(x+1)}{x+1} \right| < 0.5 $:Approximate $ \sqrt{4x^2 + 1} \approx 2x $ for large $ x $, then check:\[\frac{|1|}{x+1} < 0.5 \implies |1| < 0.5(x+1)\]Solve for $ x $:\[2(x+1) > 1 \implies x > \frac{1}{0.5} - 1 = 1\]Thus, $ N = 1 $ is one possibility.

Find Critical Value of N for 蔚 = 0.1

Perform similar steps for $ \varepsilon = 0.1 $:\[\left| \frac{\sqrt{4x^2+1} - 2(x+1)}{x+1} \right| < 0.1 \implies \frac{1}{x+1} < 0.1 \implies 1 < 0.1(x+1)\]Solve for $ x $:\[10(x+1) > 1 \implies x > \frac{1}{0.1} - 1 = 9\]Thus, $ N = 9 $ works for $ \varepsilon = 0.1 $.

Conclude with Values of N

For $ \varepsilon = 0.5 $, we found $ N = 1 $; for $ \varepsilon = 0.1 $, $ N = 9 $. This satisfies the condition for both values of $ \varepsilon $ from Definition 7.

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Key Concepts

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

Epsilon-Delta Definition

Understanding the epsilon-delta definition of a limit at infinity is crucial for grasping how limits behave when a variable approaches infinity. In the context of limits, the definition helps to encapsulate the idea that as the variable moves towards infinity, the function value gets arbitrarily close to a specific limit. This requires precision, which is where epsilon (\varepsilon) comes into play.

The limit $ \lim_{x \to \infty} f(x) = L $ is defined such that for any small positive number $ \varepsilon $, there exists a number $ N $ where for all values $ x > N $, the expression $ |f(x) - L| < \varepsilon $.
We can think of $ \varepsilon $ as the desired closeness of $ f(x) $ to $ L $, whereas $ N $ is the threshold after which all function values are that close.

In our exercise, we want to show this for the limit: $ \lim_{x \to \infty} \frac{\sqrt{4x^2+1}}{x+1} = 2 $. The task is to identify the $ N $ values that satisfy this condition for specific $ \varepsilon $ values.

Asymptotic Behavior

Asymptotic behavior gives insight into the long-term behavior of functions as inputs grow large, a common task in calculus that fosters a deep understanding of limits. The function $ f(x) = \frac{\sqrt{4x^2+1}}{x+1} $ displays such behavior as $ x \to \infty $, easing the identification of the limit.

For large $ x $, the terms $ \frac{1}{x^2} $ and $ \frac{1}{x} $ shrink towards zero.
Consequently, the expression simplifies to $ \frac{\sqrt{4}}{1} $, which equals 2, reinforcing the limit as $ x \to \infty $.

Recognizing this simplification is key as it predicts behavior at infinity, allowing us to mathematically illustrate and verify the limit value. Calculating the asymptotic form removes complexity, making further $ \varepsilon $-based calculations more straightforward and reliable.

Finding N Values

Finding the correct $ N $ values relates directly to the epsilon-delta definition and determines when $ f(x) $ gets as close to the limit as needed.
First, let's consider $ \varepsilon = 0.5 $.

To satisfy $ |f(x) - L| < 0.5 $, we approximate $ \sqrt{4x^2+1} \approx 2x $.
Solving $ \frac{|1|}{x+1} < 0.5 $ leads to $ N=1 $, implying beyond this $ N $, the function remains within 0.5 units of 2.

For $ \varepsilon = 0.1 $:

We solve $ \frac{|1|}{x+1} < 0.1 $ resulting in $ N=9 $.
This $ N $ assures that the function stays within 0.1 of 2, which illustrates a finer level of proximity.

These calculations show how to find valid $ N $ values for any $ \varepsilon $, ensuring precision in how close $ f(x) $ remains to its limit as $ x $ increases.

91影视

For the limit $$\lim _{x \rightarrow \infty} \frac{\sqrt{4 x^{2}+1}}{x+1}=2$$ illustrate Definition 7 by finding values of \(N\) that correspond to \(\varepsilon=0.5\) and \(\varepsilon=0.1\)

Short Answer

Step by step solution

Understand the Definition of a Limit

Simplify the Expression for f(x)

Establish the Inequality |f(x) - L| < 蔚

Find Critical Value of N for 蔚 = 0.5

Find Critical Value of N for 蔚 = 0.1

Conclude with Values of N

Key Concepts

Epsilon-Delta Definition

Asymptotic Behavior

Finding N Values

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