Maximizing Limit Values for Exponential Functions with Variable Parameters

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Discussion Overview

The discussion centers on evaluating the limit of the expression $\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{a}{x}\right)^{bx}$, where $a$, $b$, and $x$ are variables. Participants explore the implications of having multiple parameters and how they affect the limit, considering both theoretical and mathematical reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion regarding the presence of three variables ($a$, $b$, and $x$) in the limit expression.
  • It is noted by one participant that $a$ and $b$ are constants, and they reference known limits such as $\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{1}{x}\right)^x = e$ and $\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{a}{x}\right)^x = e^a$.
  • Another participant questions whether the constants $a$ and $b$ can be treated as scalars in the limit expression and discusses the possibility of applying L'Hôpital's rule.
  • A method involving the exponential function and logarithms is suggested, leading to the expression $\exp\left(\frac{\log\left(1+\frac{a}{x}\right)}{\frac{1}{bx}}\right)$, with the expectation that this will simplify to $e^{ab}$.
  • Clarification is provided regarding the notation "exp," equating it to the exponential function $e^{\theta}$.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the evaluation of the limit, and there are multiple competing views regarding the treatment of the variables and the application of mathematical techniques.

Contextual Notes

There is uncertainty regarding the correct application of mathematical methods, such as L'Hôpital's rule, and the treatment of constants in the limit expression. The discussion reflects varying levels of familiarity with mathematical notation and concepts.

karush
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$\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{a}{x}\right)^{bx}$

Ok I am a little stumped already because we have 3 variables in this a,x and b

W|A returned $e^{ab}$ but not what the steps are, maybe next...

$\displaystyle bx\lim_{{x}\to{\infty}}\left(1+\dfrac{a}{x}\right)$
 
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karush said:
$\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{a}{x}\right)^{bx}$

Ok I am a little stumped already because we have 3 variables in this a,x and b

note that $a$ and $b$ represent constants, and ...

$\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{1}{x}\right)^x = e$

$\displaystyle\lim_{{x}\to{\infty}}\left(1+\dfrac{a}{x}\right)^x = e^a$

... so, what happens to the limit with the constant "$b$" thrown in as an outer exponent?
 
not sure if this is the correct way to say it
but are they not just scalarsconcering what is inside the () can that be distributed or added together
 
karush said:
not sure if this is the correct way to say it
but are they not just scalars

The limit is with respect to $x$, so $a$ and $b$ are treated as constants.

Getting things into a form where we can apply L'Hopital's rule,



$$\]\exp\left(\frac{\log\left(1+\frac ax\right)}{\frac{1}{bx}}\right)$$



Now *differentiate and simplify* inside the brackets; you'll end up with $e^{ab}$.
 
Greg said:
$$\]\exp\left(\frac{\log\left(1+\frac ax\right)}{\frac{1}{bx}}\right)$$

ok I'm not familiar with "exp"
 
karush said:
ok I'm not familiar with "exp"
exp is the same as e:
[math]exp( \theta ) = e^{ \theta }[/math]

-Dan
 
made my day :cool:
 

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