Problem 64
Question
We know the integral \(\int_{1}^{x} t^{h-1} d t=\left[t^{h} / h\right]_{1}^{x}=\left(x^{h}-1\right) / h\) Its limit as \(h \rightarrow 0\) is__________.
Step-by-Step Solution
Verified Answer
The limit is \(\ln(x)\).
1Step 1: Understand the Problem
We are given the integral \(\int_{1}^{x} t^{h-1} dt = \left(t^h / h \right)_{1}^{x} = \frac{x^h - 1}{h}\). We need to find its limit as \(h \rightarrow 0\).
2Step 2: Apply the Limit
We have the expression \(\frac{x^h - 1}{h}\) and we need to find \(\lim_{h \to 0} \frac{x^h - 1}{h}\).
3Step 3: Recall Logarithmic Limit
We know from calculus, the limit \(\lim_{h \to 0} \frac{x^h - 1}{h} = \ln(x)\). This is a standard result where you recognize the definition of the derivative of \(x^h\) at \(h = 0\), yielding the natural logarithm.
Key Concepts
Limit EvaluationCalculusLogarithmic Limit
Limit Evaluation
When dealing with integration and limits, it is vital to evaluate how certain functions behave as variables approach specific values. In this case, we analyze the integral \(\int_{1}^{x} t^{h-1} dt\) as \(h\) approaches 0. Understanding the concept of limit evaluation helps you predict the behavior and outcomes of expressions under this transition.
To evaluate the limit \(\lim_{h \to 0} \frac{x^h - 1}{h}\), we consider how the expression changes as \(h\) gets closer to zero. This involves calculating the value that \(\frac{x^h - 1}{h}\) converges to, simplifying complex expressions, and understanding its defined form. Limit evaluation often requires recognizing patterns and standard results, like a known derivative or logarithmic identity, to successfully predict outcomes.
To evaluate the limit \(\lim_{h \to 0} \frac{x^h - 1}{h}\), we consider how the expression changes as \(h\) gets closer to zero. This involves calculating the value that \(\frac{x^h - 1}{h}\) converges to, simplifying complex expressions, and understanding its defined form. Limit evaluation often requires recognizing patterns and standard results, like a known derivative or logarithmic identity, to successfully predict outcomes.
Calculus
Calculus is a field in mathematics that studies change and motion. It includes the concepts of derivatives, integrals, and limits — all crucial tools for scientists and engineers. In the problem, calculus helps address the challenge of performing integration and evaluating limits to solve complex expressions.
By applying calculus principles, we integrate \(t^{h-1}\) from 1 to \(x\), resulting in the expression \(\left(t^h / h \right)_{1}^{x} = \frac{x^h - 1}{h}\). Calculus not only helps you find the definite integral of a function over an interval but also provides techniques for limit evaluation when variables tend towards certain values. Utilizing calculus techniques like these promotes deeper understanding and easier computation when dealing with diverse mathematical problems.
By applying calculus principles, we integrate \(t^{h-1}\) from 1 to \(x\), resulting in the expression \(\left(t^h / h \right)_{1}^{x} = \frac{x^h - 1}{h}\). Calculus not only helps you find the definite integral of a function over an interval but also provides techniques for limit evaluation when variables tend towards certain values. Utilizing calculus techniques like these promotes deeper understanding and easier computation when dealing with diverse mathematical problems.
Logarithmic Limit
The concept of a logarithmic limit is a crucial result in calculus, often arising during derivative calculations. It strengthens your ability to solve certain limits involving exponential functions. Here, we deal with the classic limit:\
This result is fundamental for understanding how changes in exponents can relate to logarithms. It highlights the intimate relationship between exponential growth and logarithmic functions in calculus. Grasping this concept allows you to see how logarithms serve as inverses to exponentials and provide crucial insights into the nature of change and growth in mathematical functions.
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- \(\lim_{h \to 0} \frac{x^h - 1}{h} = \ln(x)\) \
This result is fundamental for understanding how changes in exponents can relate to logarithms. It highlights the intimate relationship between exponential growth and logarithmic functions in calculus. Grasping this concept allows you to see how logarithms serve as inverses to exponentials and provide crucial insights into the nature of change and growth in mathematical functions.
Other exercises in this chapter
Problem 62
Approach a steady state \(y_{x}\) as \(t \rightarrow \infty\). True or false, when \(y_{1}\) and \(y_{2}\) both satisfy \(y^{\prime}=c y+s\) (a) The sum \(y=y_{
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Find the limit of \(\ln x / \log _{10} x\) as \(x \rightarrow \infty\).
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Find linear approximations near \(x=0\) for \(e^{-x}\) and \(2^{x}\).
View solution Problem 66
The \(x^{3}\) correction to \(\ln (1+x)\) yields \(x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}\). Check that \(\ln 1.01 \approx .0099503\) and find \(\ln 1.02\)
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