L-Hospital's Rule: Limit Calculation

[karush]

L-Hopital's rule

$\tiny{4b}$
$$\displaystyle
\begin{align}
L_{4b}&
=\lim_{{x}\to{\infty}}\left(\frac{5x+3}{5x-1}\right)^{2 x}
=e^{\frac{8}{5}}\\
\textbf{steps} \\
\ln\left({L}\right)
& =\lim_{{x}\to{\infty}}(2x)\ln\left[{\frac{5x+3}{5x-1}}\right]
=\lim_{{x}\to{\infty}} \frac{2\ln{(5x+3)}-2\ln{(5x-1)}}{{x}^{-1}} \\
\end{align} \\
\textbf{next?}$$

 

[Evgeny.Makarov]

What prevents you from applying L'Hopital's rule again?

 

[karush]

Re: L-Hopital's rule

$\tiny{4b}$
$$\displaystyle
\begin{align}%-----------------------------------------------------
L_{4b}&
=\lim_{{x}\to{\infty}}\left(\frac{5x+3}{5x-1}\right)^{2 x}
=e^{\frac{8}{5}}\\
\textit{Multiply both sides by $\ln\left({x}\right)$} \\
\ln\left({L_{4b}}\right)
& =\lim_{{x}\to{\infty}}(2x)\ln\left[{\frac{5x+3}{5x-1}}\right]
=\lim_{{x}\to{\infty}} \frac{2\ln{(5x+3)}-2\ln{(5x-1)}}{{x}^{-1}} \\
\textit{L'Hopital's Rule} \\
\ln'\left({L_{4b}}\right)&=\lim_{{x}\to{\infty}}\frac{40{x}^{2}}{(5x-1)(5x+3)}
=\lim_{{x}\to{\infty}}\left[\frac{18}{5(5x-1)}+\frac{2}{5(5x-1)}+\frac{8}{5}\right] \\
\textit{${x}\to{\infty}$} \\
\ln\left({L_{4b}}\right)& =\frac{8}{5} \\
\textit{e both sides} \\
e^{\ln\left({L_{4b}}\right)}& =L_{4b}=e^{\frac{8}{5}}
\end{align} \\ %---------------------------------------------------
$$
$\text{suggestions?}$

 

[Greg]

Multiply both sides by ln(x)? Don't you mean take the natural log of both sides?

When you reach

$$\lim_{x\to\infty}\frac{40x^2}{(5x-1)(5x+3)}$$

expand the denominator and take the limit. ;)

$$\frac{40}{25}=\frac85$$

Assuming we don't have to use L'Hopital's rule and that we're given the identity

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

we may proceed as follows:

$$\lim_{x\to\infty}\left(\frac{5x+3}{5x-1}\right)^{2x}=\lim_{x\to\infty}\left(1+\frac{4}{5x-1}\right)^{2x}$$

$$u=5x-1,\quad x=\frac{u+1}{5}$$

$$\lim_{u\to\infty}\left(1+\frac4u\right)^{2(u+1)/5}=\left(\lim_{u\to\infty}\left(1+\frac4u\right)^{u+1}\right)^{2/5}$$

$$=\left(e^4\right)^{2/5}=e^{8/5}$$

 

[karush]

Mega Mahalo..

I didn't know about the identity

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

The worksheet said the answer was $e^{20}$ ?

 

[Evgeny.Makarov]

karush, in post #3 your formulas extend past the right edge of my window. It's better to start them further to the left by not having text like "Multiply both sides..." left of the = sign or by breaking the align environment into several parts.

I didn't know about the identity

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

The worksheet said the answer was $e^{20}$ ?

The answer to what: the problem in post #1? Then why did you write from the start that the answer was $e^{8/5}$ even before you knew how to finish the solution? Or do you mean the answer to \(\displaystyle \lim_{x\to\infty}\left(1+\frac ax\right)^x\)? Then how can this be if this formula uses an unknown $a$?

 

[karush]

I generally always post the answer in the OP, mostly the book answer or from the TI Nspire CAS. the handout example gave $e^{20}$
$a$ can be determined from the given as was shown

 

[Evgeny.Makarov]

Sorry, I don't understand what you are saying.

I generally always post the answer in the OP, mostly the book answer or from the TI Nspire CAS. the handout example gave $e^{20}$

So where did the answer $e^{8/5}$ to the original problem come from: the book, TI Nspire or the handout? Is there a problem because there are two different answers: $e^{8/5}$ and $e^{20}$?

$a$ can be determined from the given

What do you mean by given?

as was shown

Shown where?

 

[karush]

it was shown in post 4
given is a common word meaning the original information
there were to answers one from handout and one from calculator which was the same as the one derived here
also when $$ is used it centers the text

 

[Evgeny.Makarov]

I am just saying that writing your thoughts more explicitly and answering all questions addressed to you would help the readers.

also when $$ is used it centers the text

That's not true, at least in my browser.

$$
\text{Equation: } (x+1)^2=x^2+2x+1
$$

gives
$$
\text{Equation: } (x+1)^2=x^2+2x+1
$$
which is closer to the left edge.

 

[karush]

there is a \begin{align} \end{align} in the lateX which is commonly used here.
I mostly use tablets I did not see clipping

 

[Greg]

To prove

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

try using L'Hopital's rule.

 

[MarkFL]

If I was going to work this problem, here's what I'd do:

\(\displaystyle L=\lim_{x\to\infty}\left(\left(\frac{5x+3}{5x-1}\right)^{2x}\right)\)

We have the indeterminate form $1^{\infty}$, and so I would take the natural log of both sides and end up with:

\(\displaystyle \ln(L)=2\lim_{x\to\infty}\left(\frac{\ln\left(\dfrac{5x+3}{5x-1}\right)}{\dfrac{1}{x}}\right)\)

Now we have the indeterminate for $\dfrac{0}{0}$, and so we may use L'Hôpital's Rule:

\(\displaystyle \ln(L)=2\lim_{x\to\infty}\left(\frac{20x^2}{25x^2+10x-3}\right)=\frac{8}{5}\)

Hence:

\(\displaystyle L=e^{\frac{8}{5}}\)