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answer the choice question upload in the picture, you only need to give a answer...
May 13, 2024
answer the choice question upload in the picture, you only need to give a answer in a,b,c or d
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Solution by Steps
step 2
The function has a vertical asymptote at x=1x = 1 because the term 1x−1\frac{1}{x - 1} becomes undefined at x=1x = 1
step 3
To find the horizontal asymptote, we examine the limit of y=xln⁡(e+1x−1)y = x \ln\left(e + \frac{1}{x - 1}\right) as xx approaches infinity
step 4
The limit can be simplified to lim⁡x→∞xln⁡(e+1x−1)=lim⁡x→∞xln⁡(x−1+ex−1)\lim_{x \to \infty} x \ln\left(e + \frac{1}{x - 1}\right) = \lim_{x \to \infty} x \ln\left(\frac{x - 1 + e}{x - 1}\right) since ee is a constant
step 5
As xx approaches infinity, the fraction x−1+ex−1\frac{x - 1 + e}{x - 1} approaches 1, and the natural logarithm of 1 is 0. Therefore, the horizontal asymptote is y=0y = 0, which is the x-axis
step 6
However, the asksia-ll calculator indicates that the function is asymptotic to y=x+1ey = x + \frac{1}{e}. This suggests that there is a slant asymptote, which occurs when the degree of the numerator is one higher than the degree of the denominator in a rational function. In this case, the function is not rational, but the behavior of the logarithm as xx approaches infinity gives a linear term xx and a constant term 1e\frac{1}{e}
B
Key Concept
Asymptotes of a function
Explanation
Asymptotes can be vertical, horizontal, or slant (oblique). Vertical asymptotes occur at points of discontinuity. Horizontal asymptotes are found by evaluating the limit of a function as xx approaches infinity. Slant asymptotes occur when the degree of the numerator is one higher than the degree of the denominator in a rational function, or through similar behavior in other types of functions.
Generate me a similar question
What is the formula for finding the volume of a cylinder
Given a_n < b_n, if sum{a_i} and sum{b_i} both converges, then sum{a_i} absolutely converge is the _ condition of sum{b_i} absolutely converge? Fill in the blank with the following options: A:sufficient but not neccesary B:sufficient and neccesary C neccesary but not sufficient D:none above
Generated Graph
Solution by Steps
step 1
To determine the relationship between the absolute convergence of ∑ai\sum{a_i} and ∑bi\sum{b_i}, we need to understand the definitions of absolute convergence
step 2
A series ∑ai\sum{a_i} is said to absolutely converge if the series ∑∣ai∣\sum{|a_i|} converges
step 3
If ∑ai\sum{a_i} converges absolutely, then ∑ai\sum{a_i} converges. However, the converse is not necessarily true
step 4
Given that a_n < b_n and both ∑ai\sum{a_i} and ∑bi\sum{b_i} converge, if ∑bi\sum{b_i} converges absolutely, then so does ∑ai\sum{a_i} because the terms of ∑ai\sum{a_i} are bounded by the terms of ∑bi\sum{b_i}
step 5
However, the absolute convergence of ∑ai\sum{a_i} does not imply the absolute convergence of ∑bi\sum{b_i}, as the terms of ∑bi\sum{b_i} could be larger in absolute value and still cause divergence
Answer
A: sufficient but not necessary
Key Concept
Absolute Convergence
Explanation
The absolute convergence of a series ∑ai\sum{a_i} where a_n < b_n is a sufficient condition for the absolute convergence of the series ∑bi\sum{b_i}, but it is not necessary. This means that if ∑bi\sum{b_i} converges absolutely, then ∑ai\sum{a_i} will also converge absolutely, but the converse may not hold
No, we can infer a_n's absolute convergence from b_n's absolute convergence This is because b_n - a_n is positive and its sum is convergence, thus absolute convergence, and we can view a_n = b_n +n (a_n - b_n), the sum of two absolute convergence series also convergence
Solution by Steps
step 1
Evaluate the limit of the given expression as nn approaches infinity
step 2
Simplify the expression inside the limit: ln⁡(n+1)3nn3ln⁡(n)3n+1(n+1)3\frac{\ln(n+1)3^n n^3}{\ln(n)3^{n+1}(n+1)^3}
step 3
Factor out the common terms and apply the properties of logarithms: ln⁡(n+1)ln⁡(n)⋅3n3n+1⋅n3(n+1)3\frac{\ln(n+1)}{\ln(n)} \cdot \frac{3^n}{3^{n+1}} \cdot \frac{n^3}{(n+1)^3}
step 4
Simplify the expression further: ln⁡(n+1)ln⁡(n)⋅13⋅(nn+1)3\frac{\ln(n+1)}{\ln(n)} \cdot \frac{1}{3} \cdot \left(\frac{n}{n+1}\right)^3
step 5
Take the limit of each part separately as nn approaches infinity: lim⁡n→∞ln⁡(n+1)ln⁡(n)=1\lim_{n\to\infty} \frac{\ln(n+1)}{\ln(n)} = 1, lim⁡n→∞13=13\lim_{n\to\infty} \frac{1}{3} = \frac{1}{3}, and lim⁡n→∞(nn+1)3=1\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^3 = 1
step 6
Combine the results of the limits: 1⋅13⋅1=131 \cdot \frac{1}{3} \cdot 1 = \frac{1}{3}
step 7
Since we are dealing with the absolute value, the final result is the absolute value of 13\frac{1}{3}, which is 13\frac{1}{3}
Answer
The limit of the given expression as nn approaches infinity is 13\frac{1}{3}.
Key Concept
Limits of functions involving logarithms and exponential terms as nn approaches infinity
Explanation
The key concept in finding the limit of the given expression is to simplify the terms and take the limit of each part separately. The properties of logarithms and the behavior of exponential functions as nn approaches infinity are used to determine the limit.
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