Chapter 5: Problem 7
Argue that $$\lim _{b \rightarrow a} \frac{e^{\sqrt{b}}-e^{\sqrt{a}}}{\sqrt{b}-\sqrt{a}}=e^{\sqrt{a}}$$
Short Answer
Expert verified
The limit evaluates to \( e^{\sqrt{a}} \) by using the derivative of the function \( f(x) = e^{\sqrt{x}} \) at \( x = a \).
Step by step solution
01
Understanding the Problem
We need to evaluate the limit \( \lim_{b \rightarrow a} \frac{e^{\sqrt{b}} - e^{\sqrt{a}}}{\sqrt{b} - \sqrt{a}} \) and show that it equals \( e^{\sqrt{a}} \). This type of problem is a derivative at a point problem, relating to the derivative definition of a function using limits.
02
Recognize the Limit as a Derivative
Recognize that the given limit expression resembles the definition of a derivative: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). Thus, observe that \( \lim_{b \rightarrow a} \frac{e^{\sqrt{b}} - e^{\sqrt{a}}}{\sqrt{b} - \sqrt{a}} \) is effectively the derivative of the function \( f(x) = e^{\sqrt{x}} \) evaluated at \( x = a \).
03
Find the Derivative using the Chain Rule
Let's find \( f'(x) \) for the function \( f(x) = e^{\sqrt{x}} \). Use the chain rule: if \( f(x) = e^{g(x)} \), then \( f'(x) = e^{g(x)} \cdot g'(x) \). In our case, \( g(x) = \sqrt{x} \), so \( g'(x) = \frac{1}{2\sqrt{x}} \). Therefore, \( f'(x) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \).
04
Evaluate the Derivative at the Point
The derivative we found is \( f'(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \). Evaluating this at \( x = a \), we get \( f'(a) = \frac{e^{\sqrt{a}}}{2\sqrt{a}} \cdot 2\sqrt{a} = e^{\sqrt{a}} \). Thus, the limit equals \( e^{\sqrt{a}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
Understanding derivatives is fundamental to calculus. A derivative represents the rate at which a function is changing at any given point.
In simple terms, it tells us how steep a curve is at a particular point on the graph of a function. Mathematically, the derivative of a function, usually denoted as \( f'(x) \), is defined as the limit:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
This limit expression calculates the slope of the tangent line to the curve at the point \( x \).
When solving problems involving derivatives using limits, we often express incremental changes in the function's inputs and outputs, just like the problem discussed where we calculated a limit at \( b \rightarrow a \) using the derivative definition.
In simple terms, it tells us how steep a curve is at a particular point on the graph of a function. Mathematically, the derivative of a function, usually denoted as \( f'(x) \), is defined as the limit:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
This limit expression calculates the slope of the tangent line to the curve at the point \( x \).
When solving problems involving derivatives using limits, we often express incremental changes in the function's inputs and outputs, just like the problem discussed where we calculated a limit at \( b \rightarrow a \) using the derivative definition.
- Derivatives are essential in measuring how dynamic quantities change over time.
- They are used in various fields such as physics, engineering, and economics to model real-world scenarios.
Chain Rule
The chain rule is a powerful tool in calculus used to differentiate composite functions.
It allows us to find the derivative of a function that is composed of other functions.
If we have a function \( f(x) = e^{\sqrt{x}} \), we need to use the chain rule to find its derivative efficiently.
The chain rule states that if a function \( f \) can be expressed as \( f(x) = g(h(x)) \), then its derivative \( f'(x) \) is given by:
\[f'(x) = g'(h(x)) \cdot h'(x)\]
In our situation, let \( g(x) = e^x \) and \( h(x) = \sqrt{x} \). The derivative \( g'(x) = e^x \) and \( h'(x) = \frac{1}{2\sqrt{x}} \).
Therefore, using the chain rule, the derivative \( f'(x) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \).
The chain rule simplifies the process of differentiating complex functions by breaking them down into simpler parts.
It allows us to find the derivative of a function that is composed of other functions.
If we have a function \( f(x) = e^{\sqrt{x}} \), we need to use the chain rule to find its derivative efficiently.
The chain rule states that if a function \( f \) can be expressed as \( f(x) = g(h(x)) \), then its derivative \( f'(x) \) is given by:
\[f'(x) = g'(h(x)) \cdot h'(x)\]
In our situation, let \( g(x) = e^x \) and \( h(x) = \sqrt{x} \). The derivative \( g'(x) = e^x \) and \( h'(x) = \frac{1}{2\sqrt{x}} \).
Therefore, using the chain rule, the derivative \( f'(x) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \).
The chain rule simplifies the process of differentiating complex functions by breaking them down into simpler parts.
- It is used to solve problems that involve more than one variable or function.
- This rule expands our ability to work with a wider variety of functions in calculus.
Function Limit
The concept of a function limit is central to calculus, especially when working with derivatives and integrals.
A function limit describes what value a function approaches as the input approaches a certain point.
The limit that we evaluated in this problem was:
\[\lim_{b \rightarrow a} \frac{e^{\sqrt{b}} - e^{\sqrt{a}}}{\sqrt{b} - \sqrt{a}}\]
This limit was solved by recognizing it as a derivative in disguise. It tells us how \( e^{\sqrt{x}} \) changes as \( x \) approaches \( a \).
Limits allow us to deal with quantities that are not well-defined at a certain point directly but can be approximated as closely as desired.
A function limit describes what value a function approaches as the input approaches a certain point.
The limit that we evaluated in this problem was:
\[\lim_{b \rightarrow a} \frac{e^{\sqrt{b}} - e^{\sqrt{a}}}{\sqrt{b} - \sqrt{a}}\]
This limit was solved by recognizing it as a derivative in disguise. It tells us how \( e^{\sqrt{x}} \) changes as \( x \) approaches \( a \).
- Function limits are crucial when defining derivatives and integrations.
- They help in understanding the behavior of functions near specific points.
Limits allow us to deal with quantities that are not well-defined at a certain point directly but can be approximated as closely as desired.
Mathematical Proofs
Mathematical proofs are logical arguments demonstrating the truth of a mathematical statement.
They are essential for validating results and establishing certainty in mathematics.
In our problem, proving that a derivative equals the limit as \( b \rightarrow a \) is a form of proof.
Proofs in calculus, especially involving limits and derivatives, require a careful application of definitions and properties.
They are essential for validating results and establishing certainty in mathematics.
In our problem, proving that a derivative equals the limit as \( b \rightarrow a \) is a form of proof.
Proofs in calculus, especially involving limits and derivatives, require a careful application of definitions and properties.
- They rely on understanding precise definitions, such as the limit definition of a derivative.
- Proofs can vary from direct proofs, using straightforward logical deduction, to indirect proofs, employing methods like contradiction or contraposition.