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Use Definition \(1.4 .1\) to prove that the stated limit is correct. In each case, to show that \(\lim _{x \rightarrow a} f(x)=L\), factor \(|f(x)-L|\) in the form $$ |f(x)-L|=\mid \text { "something"| }|\cdot| x-a \mid $$ and then bound the size of \(\mid\) "something"l by putting restrictions on the size of \(\delta\). $$ \lim _{x \rightarrow 4} \sqrt{x}=2 $$

Short Answer

Expert verified
The limit \( \lim_{x \rightarrow 4} \sqrt{x} = 2 \) is proven by showing \(|\sqrt{x} - 2| = \frac{|x - 4|}{|\sqrt{x} + 2|}\) and bounding \(|\sqrt{x} + 2|\).

Step by step solution

01

Understand Definition 1.4.1 and the Problem

Definition 1.4.1 states that \( \lim_{x \rightarrow a} f(x) = L \) if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x - a| < \delta \), it follows that \( |f(x) - L| < \epsilon \). To solve this exercise, we need to show \( \lim _{x \rightarrow 4} \sqrt{x} = 2 \) by factoring the absolute difference \( |f(x) - L| \) involving \( x - a \).
02

Express the Difference |f(x) - L|

Our function is \( \sqrt{x} \) and we want to prove \( \lim_{x \rightarrow 4} \sqrt{x} = 2 \). Evaluate the expression \(|f(x) - L| = |\sqrt{x} - 2|\) and factor it in terms of \( |x - a| \) where \( a = 4 \).
03

Factor |f(x) - L|

Using the identity for differences between roots, we get:\[|\sqrt{x} - 2| = |\sqrt{x} - \sqrt{4}| = \frac{|x - 4|}{|\sqrt{x} + 2|}.\] Here, \(|\sqrt{x} + 2|\) is the bounded factor for suitable values of \(x\).
04

Bound the Expression

To provide a bound for \(|\sqrt{x} + 2|\), assume \(x\) is close to 4. For example, if \(0 < |x - 4| < 1\), then \(3 < x < 5\).In this range, \(\sqrt{x}\) lies between \(\sqrt{3}\) and \(\sqrt{5}\). Therefore, \(\sqrt{x} + 2\) lies between approximately 3.73 and 4.24, so \[\frac{1}{|\sqrt{x} + 2|} < 0.27.\]
05

Establish the Delta (\(\delta\)) Condition

For any \(\epsilon > 0\), choose \( \delta = \min(1, \epsilon \times 0.27) \). Then whenever \(0 < |x - 4| < \delta\):\[|\sqrt{x} - 2| = \frac{|x - 4|}{|\sqrt{x} + 2|} < \delta \cdot 0.27 \leq \epsilon.\] Hence, \(|\sqrt{x} - 2| < \epsilon\), verifying the limit statement using the definition of a limit.

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Key Concepts

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

Epsilon-Delta Definition
Understanding limits through the epsilon-delta definition is a cornerstone of calculus.
This definition gives a rigorous way to say that a function approaches a certain value as the input gets close to a particular point. It formally states that \( \lim_{x \rightarrow a} f(x) = L \) if for every \( \epsilon > 0 \) (where \( \epsilon \) is any small positive number), there exists a corresponding \( \delta > 0 \) so that whenever the distance \( |x - a| \) is less than \( \delta \) but greater than zero, the distance \( |f(x) - L| \) is less than \( \epsilon \).
In simple terms:
  • \( \epsilon \) controls how close \( f(x) \) needs to be to \( L \).
  • \( \delta \) dictates how close \( x \) should be to \( a \) for \( f(x) \) to stay within that range of closeness defined by \( \epsilon \).
This concept is pivotal in proving limits because it lets us mathematically ensure that a function behaves as expected in the neighborhood of a point.
Factoring Expressions
Factoring expressions, especially differences of functions, is a valuable skill in calculus when dealing with limits.
When tasked with finding \( \lim_{x \rightarrow a} f(x) = L \), we often need to express the difference \( |f(x) - L| \) in a form that highlights the role of \( |x-a| \).
This usually involves rewriting \( |f(x) - L| \) to explicitly show the product \( |s(x)| \cdot |x-a| \), where \( s(x) \) is a new expression that doesn’t vanish as \( x \) approaches \( a \).In the given problem, \( f(x) = \sqrt{x} \) and \( L = 2 \).
We factor it like:
  • Recognize \( |\sqrt{x} - L| = |\sqrt{x} - 2| \) as a classic difference of square roots.
  • Using the identity \( |\sqrt{x} - 2| = \frac{|x - 4|}{|\sqrt{x} + 2|} \), we can uncover the structure around \( x - 4 \).
This factoring helps isolate the part influenced by \( x \'s \) proximity to 4, making it easier to apply the epsilon-delta definition.
Limits of Square Roots
Calculating the limits of square root functions involves careful handling of their algebraic properties.
Square roots often complicate the process because they introduce non-linear behavior that isn't as easily handled algebraically compared to polynomials.
Nevertheless, factoring and bounding techniques simplify the analysis.For \( \lim_{x \rightarrow 4} \sqrt{x} = 2 \), we proceed by:
  • Factoring the difference \( |\sqrt{x} - 2| \) into \( \frac{|x-4|}{|\sqrt{x} + 2|} \).
  • Bounding \( |\sqrt{x} + 2| \) to make it easier to find a suitable \( \delta \).
We choose the bound by considering the behavior of \( \sqrt{x} \) in a small open interval around the point \( x = 4 \).
By bounding \( \frac{1}{|\sqrt{x} + 2|} \), we ensure that the entire expression \( |\sqrt{x} - 2| \) becomes small as \( x \) gets close to 4.
This method demonstrates effectively verifying limits for square root functions using algebraic manipulation and bounding insights.

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