- #1
gayani
- 2
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Homework Statement
Find A so that lim (x+A/x-2A)^x =5
x-> 00
Homework Equations
The Attempt at a Solution
got to use lopihtals rule.
How to Use L'Hôpital's Rule to Find A in a Limit Equation
- Thread starter gayani
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Homework Help Overview
The problem involves finding a value for A such that the limit of the expression \((x + A)/(x - 2A)\) raised to the power of x approaches 5 as x approaches infinity. The context is centered around the application of L'Hôpital's rule and limit evaluation.
Discussion Character
- Exploratory, Mathematical reasoning, Assumption checking
Approaches and Questions Raised
- Participants discuss the use of L'Hôpital's rule and suggest rewriting the limit expression in a different form to facilitate analysis. There is also clarification regarding the interpretation of the original expression, with some participants questioning the correct setup of the limit.
Discussion Status
The discussion is ongoing, with participants providing guidance on how to manipulate the expression for limit evaluation. There is an emphasis on understanding the structure of the limit and the application of logarithmic properties to assist in the analysis.
Contextual Notes
There is some confusion regarding the initial expression's interpretation, which may affect the approach to finding A. Participants are exploring different interpretations and methods without reaching a consensus yet.
Find A so that lim (x+A/x-2A)^x =5
x-> 00
got to use lopihtals rule.
- #2
tiny-tim
Science Advisor
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Welcome to PF!
Hi gayani! Welcome to PF!
(have an infinity: ∞ and try using the X2 tag just above the Reply box
)
hmm … I can't see how to use l'Hôpital's rule
…
but if you write the LHS as (1 + 3A/(x - 2A))x, and fiddle about with it, you should get something you recognise.
Hi gayani! Welcome to PF!
(have an infinity: ∞ and try using the X2 tag just above the Reply box
)gayani said:
hmm … I can't see how to use l'Hôpital's rule
…but if you write the LHS as (1 + 3A/(x - 2A))x, and fiddle about with it, you should get something you recognise.
- #3
Mark44
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Gayani,
Just in case you don't follow what Tiny-tim did, here it is.
You wrote the rational expression as x + A/x - 2A. What I'm pretty sure you meant (and the way Tiny-tim interpreted it) was (x + A)/(x - 2A). What you wrote would be correctly interpreted as the sum and difference of three expressions-- x, A/x, and 2/A, not the quotient of x + A and x - 2A.
What Tiny-tim did was divide x + A by x - 2A, so that (x + A)/(x - 2A) = 1 + 3A/(x - 2A).
Just in case you don't follow what Tiny-tim did, here it is.
You wrote the rational expression as x + A/x - 2A. What I'm pretty sure you meant (and the way Tiny-tim interpreted it) was (x + A)/(x - 2A). What you wrote would be correctly interpreted as the sum and difference of three expressions-- x, A/x, and 2/A, not the quotient of x + A and x - 2A.
What Tiny-tim did was divide x + A by x - 2A, so that (x + A)/(x - 2A) = 1 + 3A/(x - 2A).
- #4
Bohrok
- 867
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Take the natural log of both sides and then take the limit of the natual log of the left side
[tex]\lim_{x\rightarrow\infty} ln\left(1 + \frac{3A} {x-2A}\right)^x= ln (5)[/tex]
You'll have to do some rearranging in order to use l'Hôpital's rule and "solve" for A.
[tex]\lim_{x\rightarrow\infty} ln\left(1 + \frac{3A} {x-2A}\right)^x= ln (5)[/tex]
You'll have to do some rearranging in order to use l'Hôpital's rule and "solve" for A.
Last edited:
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