Maximizing Limit Values for Exponential Functions with Variable Parameters

[karush]

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

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...

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

 

[skeeter]

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

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?

 

[karush]

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

 

[Greg]

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,



\(\displaystyle \]\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 $e^{ab}$.

 

[karush]

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

ok I'm not familiar with "exp"

 

[topsquark]

ok I'm not familiar with "exp"

exp is the same as e:
\(\displaystyle exp( \theta ) = e^{ \theta }\)

-Dan

 

[karush]

made my day