Chapter 11: Problem 41
Find the limit of the sequence. $$\lim _{n \rightarrow \infty} \frac{n^{2} \ln n}{e^{n}}$$
Short Answer
Expert verified
The limit of the sequence \(\lim _{n \rightarrow \infty} \frac{n^{2} \ln n}{e^{n}}\) is \(0\).
Step by step solution
01
Verify conditions for L'Hôpital's rule
To apply L'Hôpital's rule, we must first verify the following conditions for the given sequence:
1. Both the numerator and denominator approach infinity as n approaches infinity
As n approaches infinity, \(\lim_{n \rightarrow \infty} n^2 \ln n = \infty\), and \(\lim_{n \rightarrow \infty} e^n = \infty\).
02
Apply L'Hôpital's rule
Since both the numerator and the denominator approach infinity, we can apply L'Hôpital's rule:
\(\lim_{n \rightarrow \infty} \frac{n^2 \ln n}{e^n} = \lim_{n \rightarrow \infty} \frac{\frac{d}{dn} (n^2 \ln n)}{\frac{d}{dn} (e^n)}\)
Now, we need to find the derivatives of the numerator and denominator.
03
Find the derivative of the numerator
To find the derivative of the numerator, use the product rule for differentiation:
\(\frac{d}{dn}(n^2 \ln n) = n^2 \frac{d}{dn}(\ln n) + \ln n \frac{d}{dn}(n^2)\)
\(\frac{d}{dn}(n^2 \ln n) = n^2 \frac{1}{n} + \ln n(2n)\)
\(\frac{d}{dn}(n^2 \ln n) = n + 2n \ln n\)
04
Find the derivative of the denominator
The derivative of the denominator is simply:
\(\frac{d}{dn} (e^n) = e^n\)
05
Find the limit after first L'Hôpital's rule application
Plug the derivatives back into the limit equation:
\(\lim_{n \rightarrow \infty} \frac{n + 2n \ln n}{e^n}\)
Both the numerator and the denominator approach infinity as n approaches infinity, so we need to apply L'Hôpital's rule one more time.
06
Apply L'Hôpital's rule again
\(\lim_{n \rightarrow \infty} \frac{n + 2n \ln n}{e^n} = \lim_{n \rightarrow \infty} \frac{\frac{d}{dn}(n + 2n \ln n)}{\frac{d}{dn}(e^n)}\)
Find the derivative of the new numerator and denominator:
07
Find the derivative of the new numerator
\(\frac{d}{dn}(n + 2n \ln n) = 1 + 2(\ln n + 1)\)
08
Find the derivative of the new denominator again
The derivative of the denominator remains the same:
\(\frac{d}{dn} (e^n) = e^n\)
09
Find the limit after the second L'Hôpital's rule application
Plug the derivatives back into the limit equation:
\(\lim_{n \rightarrow \infty} \frac{1 + 2(\ln n + 1)}{e^n}\)
Now, the numerator approaches a constant as n approaches infinity, while the denominator approaches infinity:
\(\lim_{n \rightarrow \infty} \frac{1 + 2(\ln n + 1)}{e^n} = 0\)
So, the limit of the given sequence is \(0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus for evaluating limits, especially when direct substitution into a limit results in an indeterminate form such as \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\). To apply L'Hôpital's Rule, you must first ensure that the conditions for use are satisfied:
- Both the numerator and denominator approach 0 or both approach infinity.
- The functions are differentiable in a neighborhood around the point of interest.
Limit of a Sequence
The limit of a sequence \((a_n)\) as \(n\) approaches infinity is a central theme in calculus and analysis. It captures the idea of a sequence approaching a particular value. If a sequence \((a_n)\) approaches \(L\) as \(n\) goes to infinity, we write:\[\lim_{n \to \infty} a_n = L\]For our solution, we determined that the limit of the sequence \(\lim_{n \to \infty} \frac{n^2 \ln n}{e^n}\) is 0. This indicates that as \(n\) becomes very large, the values of the sequence get closer and closer to 0.
- This concept is especially useful when investigating long-term behavior of sequences in fields like economics, physics, and computer science.
Differentiation
Differentiation is a fundamental aspect of calculus, focusing on calculating the derivative of a function. Derivatives represent rates of change and are critical in numerous real-world applications. When tackling our problem, we utilized differentiation multiple times as part of L'Hôpital's Rule to find the derivatives of the numerator \(n^2 \ln n\) and the denominator \(e^n\).
- The product rule allowed us to differentiate \(n^2 \ln n\), which is the product of \(n^2\) and \(\ln n\).
- The exponential function \(e^n\), known for its simple derivative, made calculations straightforward.
Exponential Functions
Exponential functions are a class of functions where the variable appears in the exponent, typically in the form \(e^x\). They exhibit rapid growth or decay characteristics, which significantly influence the behavior of sequences and functions in calculus.
- The base of the natural logarithm \(e\) is approximately 2.718 and is unique in that the function \(e^x\) is its own derivative.
- In the limit \(\lim_{n \to \infty} \frac{n^2 \ln n}{e^n}\), the exponential function in the denominator grows much faster than any polynomial or logarithmic terms in the numerator, particularly as \(n\) approaches infinity.