Problem 31
Question
\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow \infty}(x-\ln x)$$
Step-by-Step Solution
Verified Answer
The limit is infinity.
1Step 1: Identify the expression
We are given the expression \(x - \ln x\). We need to find the limit as \(x\) approaches infinity, i.e., \(\lim_{x \to \infty} (x - \ln x)\).
2Step 2: Analyze the terms separately
As \(x\to\infty\), the term \(x\) in the expression \(x - \ln x\) tends to infinity. The natural logarithm term, \(\ln x\), also tends to infinity but does so at a slower rate compared to \(x\).
3Step 3: Use an intuitive approach
Since both \(x\) and \(\ln x\) tend to infinity, we check the rates at which these grow. The function \(x\) is linear, while \(\ln x\) increases much more slowly. Subtracting a slower-growing infinity (\(\ln x\)) from a faster-growing infinity (\(x\)) leads us to suspect the limit might approach infinity.
4Step 4: Conclude the result based on elementary considerations
Since \(x\) grows much faster than \(\ln x\), the term \(x - \ln x\) will continuously increase without bound as \(x\to\infty\). Thus, the limit is \(\infty\).
Key Concepts
Limits at InfinityNatural Logarithm FunctionRates of Growth in Calculus
Limits at Infinity
When you encounter the concept of limits at infinity, you are essentially exploring what happens to a function as the input value grows larger and larger without bound. Playing a vital role in calculus, limits at infinity help us understand the behavior of functions that continue indefinitely.
To evaluate limits at infinity, you should look at the dominant term of the function. In this case, the expression is \(x - \ln x\).
As \(x\) increases, each part of the expression behaves differently:
To evaluate limits at infinity, you should look at the dominant term of the function. In this case, the expression is \(x - \ln x\).
As \(x\) increases, each part of the expression behaves differently:
- \(x\) grows without restriction, becoming very large.
- \(\ln x\) also increases but at a slower pace since it is logarithmic.
Natural Logarithm Function
The natural logarithm function, denoted as \(\ln(x)\), is a specific logarithmic function with a base of \(e\), where \(e\) is an irrational constant approximately equal to 2.718. \(\ln(x)\) is the inverse of the exponential function \(e^x\), making it uniquely important in various mathematical contexts.
Properties of the natural logarithm include:
Properties of the natural logarithm include:
- \(\ln(1) = 0\) because \(e^0 = 1\).
- It is defined only for positive numbers, \(x > 0\).
- The graph of \(\ln(x)\) is continuous and increases monotonically, but does so more slowly than linear, polynomial, or exponential functions.
Rates of Growth in Calculus
In calculus, rates of growth refer to how rapidly a function increases or decreases as its input grows. Knowing how different functions grow relative to one another helps in predicting and analyzing their behavior over infinite intervals.
Common growth rates in functions include:
Common growth rates in functions include:
- Constant Functions: The simplest, where growth is zero. \(f(x) = c\).
- Logarithmic Functions: Like \(\ln x\), increase slowly.
- Linear Functions: Like \(x\), increase steadily and directly.
- Polynomial Functions: Typically increase faster than linear functions, especially as the degree of the polynomial becomes larger.
- Exponential Functions: Grow very rapidly, far outpacing other types at infinity.
Other exercises in this chapter
Problem 31
Differentiate the function. $$ f(t)=\tan \left(e^{t}\right)+e^{\tan t} $$
View solution Problem 31
Find the derivative of the function. Find the domains of the function and its derivative. \(g(x)=\cos ^{-1}(3-2 x)\)
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(a) Show that \(f\) is one-to-one. (b) Use Theorem 7 to find \(\left(f^{-1}\right)^{\prime}(a)\) . (c) Calculate \(f^{-1}(x)\) and state the domain and range of
View solution Problem 31
If you graph the function \(f(x)=\frac{1-e^{1 / x}}{1+e^{1 / x}}\) you'll see that \(f\) appears to be an odd function. Prove it.
View solution