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Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]?
Most Upvoted Answer
Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]?
Explanation:
To find the value of the limit as n approaches infinity for the expression log((1+x)^2) - log(x)^2, we can use the properties of logarithms and the limit laws.
Step 1: Rewrite the expression
The expression can be rewritten using the logarithmic identity log(a^b) = b*log(a):
log((1+x)^2) - log(x)^2 = 2*log(1+x) - log(x)^2
Step 2: Simplify the expression
We can further simplify the expression by applying the logarithmic properties. The subtraction of logarithms can be expressed as the division of the corresponding arguments:
2*log(1+x) - log(x)^2 = log((1+x)^2/x^2)
Step 3: Apply the limit
To find the limit as n approaches infinity, we substitute x with n and evaluate the expression:
lim(n->∞) log((1+n)^2/n^2)
Step 4: Simplify the expression further
We can simplify the expression by expanding the numerator and denominator:
lim(n->∞) log(1+2n+n^2/n^2)
Step 5: Apply the limit laws
Since the logarithm function is continuous, we can apply the limit inside the logarithm:
log(lim(n->∞) (1+2n+n^2/n^2))
Step 6: Evaluate the limit
As n approaches infinity, the term (1+2n+n^2/n^2) approaches 1. Therefore, the limit becomes:
log(1) = 0
Conclusion:
The limit as n tends to infinity for the expression log((1+x)^2) - log(x)^2 is equal to 0.
To find the value of the limit as n approaches infinity for the expression log((1+x)^2) - log(x)^2, we can use the properties of logarithms and the limit laws.
Step 1: Rewrite the expression
The expression can be rewritten using the logarithmic identity log(a^b) = b*log(a):
log((1+x)^2) - log(x)^2 = 2*log(1+x) - log(x)^2
Step 2: Simplify the expression
We can further simplify the expression by applying the logarithmic properties. The subtraction of logarithms can be expressed as the division of the corresponding arguments:
2*log(1+x) - log(x)^2 = log((1+x)^2/x^2)
Step 3: Apply the limit
To find the limit as n approaches infinity, we substitute x with n and evaluate the expression:
lim(n->∞) log((1+n)^2/n^2)
Step 4: Simplify the expression further
We can simplify the expression by expanding the numerator and denominator:
lim(n->∞) log(1+2n+n^2/n^2)
Step 5: Apply the limit laws
Since the logarithm function is continuous, we can apply the limit inside the logarithm:
log(lim(n->∞) (1+2n+n^2/n^2))
Step 6: Evaluate the limit
As n approaches infinity, the term (1+2n+n^2/n^2) approaches 1. Therefore, the limit becomes:
log(1) = 0
Conclusion:
The limit as n tends to infinity for the expression log((1+x)^2) - log(x)^2 is equal to 0.
Community Answer
Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]?
Question: Evaluate the limit as n tends to infinity of [{log(1 + x)}^2 - {log(x)}^2].
Solution:
To find the limit of the given expression as n tends to infinity, we can simplify the expression and analyze its behavior.
Simplifying the expression:
Let's start by expanding the given expression:
[{log(1 + x)}^2 - {log(x)}^2]
Using the identity log(a) - log(b) = log(a/b), we can rewrite the expression as:
[{log((1 + x)/x)} * {log((1 + x)/x)}]
Simplifying further, we get:
[{log((1 + x)/(x))} * {log((1 + x)/(x))}]
Asymptotic Behavior:
To analyze the asymptotic behavior of the expression as n tends to infinity, we can consider the behavior of the individual logarithmic terms.
Logarithmic term 1: log((1 + x)/(x))
As x approaches infinity, the term (1 + x)/x approaches 1, since the x term dominates the 1 term. Therefore, we have:
log((1 + x)/(x)) ≈ log(1) = 0
Logarithmic term 2: log((1 + x)/(x))
Similarly, as x approaches infinity, the term (1 + x)/x approaches 1, and we have:
log((1 + x)/(x)) ≈ log(1) = 0
Final expression:
Substituting the asymptotic behavior of the logarithmic terms back into the original expression, we get:
[{log((1 + x)/(x))} * {log((1 + x)/(x))}] ≈ [0 * 0] = 0
Conclusion:
Therefore, as n tends to infinity, the given expression [{log(1 + x)}^2 - {log(x)}^2] approaches 0.
Solution:
To find the limit of the given expression as n tends to infinity, we can simplify the expression and analyze its behavior.
Simplifying the expression:
Let's start by expanding the given expression:
[{log(1 + x)}^2 - {log(x)}^2]
Using the identity log(a) - log(b) = log(a/b), we can rewrite the expression as:
[{log((1 + x)/x)} * {log((1 + x)/x)}]
Simplifying further, we get:
[{log((1 + x)/(x))} * {log((1 + x)/(x))}]
Asymptotic Behavior:
To analyze the asymptotic behavior of the expression as n tends to infinity, we can consider the behavior of the individual logarithmic terms.
Logarithmic term 1: log((1 + x)/(x))
As x approaches infinity, the term (1 + x)/x approaches 1, since the x term dominates the 1 term. Therefore, we have:
log((1 + x)/(x)) ≈ log(1) = 0
Logarithmic term 2: log((1 + x)/(x))
Similarly, as x approaches infinity, the term (1 + x)/x approaches 1, and we have:
log((1 + x)/(x)) ≈ log(1) = 0
Final expression:
Substituting the asymptotic behavior of the logarithmic terms back into the original expression, we get:
[{log((1 + x)/(x))} * {log((1 + x)/(x))}] ≈ [0 * 0] = 0
Conclusion:
Therefore, as n tends to infinity, the given expression [{log(1 + x)}^2 - {log(x)}^2] approaches 0.
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Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]?
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Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]?.
Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Lim n tending to infinity [{log(1 x)}^2-{log(x)}^2 ]?.
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