Error at the very end of simple limit x to inf (1 - ( 1 / ( (4x+5) / 3 ) ) )^(x-6)

[kubek]

Hello,

I've been trying to find the error that I am doing at the very end, for, at least, last 2 hours, and I think I need another pair of eyes

1. This is the simple limit that I know is true and I would like to get to that state:

. . . . .\(\displaystyle \large{\displaystyle \lim_{x \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{x}\right)^x\, =\, e}\)

2. On the left is the starting state and my first steps I did. I know these steps should be correct:

. . . . .\(\displaystyle \large{\displaystyle \lim_{x \rightarrow \infty}\, \left(\dfrac{4x\, +\, 2}{4x\, +\, 5}\right)^{x - 6}\, =\, \lim_{x \rightarrow \infty}\, \left(1\, -\, \dfrac{3}{4x\, +\, 5}\right)^{x - 6}\, =\, \lim_{x \rightarrow \infty}\, \left(1\, -\, \dfrac{1}{\left(\frac{4x\, +\, 5}{3}\right)}\right)^{x - 6}}\)

3. I did substitution as can be seen below (I maybe did some error here; three dots are just delimiters, they don't have meaning):

. . . . .\(\displaystyle \large{\displaystyle a\, =\, -\, \dfrac{(4x\, +\, 5)}{3}\, ...\, x\, =\, \dfrac{(-3a\, -\, 5)}{4}\, ...\, x\, -\, 6\, =\, \dfrac{(-3a\, -\, 29)}{4}}\)

4. Used in the expression, I tried to finish it to unsuccesful ending. In the below steps, there is almost for sure error:

. . . . .\(\displaystyle \large{\displaystyle \lim_{a\, \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{3a - 29}{4}}\, =\, \lim_{a \rightarrow \infty}\, \bigg(\left(1\, +\, \dfrac{1}{a}\right)^a\bigg)^{-\frac{26}{4}}\, =\, e^{-\frac{13}{2}}}\)

While the correct result should be:

. . . . .\(\displaystyle \large{\displaystyle e^{-\frac{3}{4}}}\)

Can anyone please see some error(s)?

I hope I was clear enough.

Thank you very much

 

[ksdhart2]

Everything you've done seems fine up until this step:

\(\displaystyle \displaystyle \lim _{a\to \infty }\left(\left(1+\dfrac{1}{a}\right)^{-\dfrac{3a-29}{4}}\right)=\lim _{a\to \infty }\left(\left(\left(1+\dfrac{1}{a}\right)^a\right)^{-\dfrac{26}{4}}\right)\)

The above is not an identity - it's only true if \(\displaystyle a \approx 1.26087\). What were your steps at arriving at the above equation?

 

[Steven G]

Everything you've done seems fine up until this step:

\(\displaystyle \displaystyle \lim _{a\to \infty }\left(\left(1+\dfrac{1}{a}\right)^{-\dfrac{3a-29}{4}}\right)=\lim _{a\to \infty }\left(\left(\left(1+\dfrac{1}{a}\right)^a\right)^{-\dfrac{26}{4}}\right)\)

The above is not an identity - it's only true if \(\displaystyle a \approx 1.26087\). What were your steps at arriving at the above equation?

(NOT at ksdhart2)----can anyone say ln???

 

[pka]

\(\displaystyle \large{\left(\dfrac{x+2}{4x+5}\right)^{x-6}=\left(1-\dfrac{3/4}{x+5/4}\right)^{x}\left(\dfrac{4x+5}{4x+2}\right)^{6} \to e^{3/4}}\)

 

[kubek]

Thank you ksdhart2 and pka. Jomo, I guess you wanted me to use natural logarithm, but I wanted to do without it, nonetheless, thank you.

So I see the error that I've done was that I haven't realized there is a substraction in my exponent. "Splitting it up" got me to the correct result:

. . . . .\(\displaystyle \large{ \displaystyle \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{3a}{4}}\, \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{29}{4}}\, =\, \bigg(\, \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^a\, \bigg)^{-\frac{3}{4}} }\)

. . . . .\(\displaystyle \large{ \displaystyle \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{29}{4}}\, =\, e^{-\frac{3}{4}}\, \cdot\, 1\, =\, e^{-\frac{3}{4}} }\)

I'm sorry for such a simple mistake, I guess this hardly belongs to the Calculus forum. Please excuse me also for not using Tex, which I'll take a look at very soon.

Thank you all very much

 

[lookagain]

. . . . .\(\displaystyle \large{ \displaystyle \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{3a}{4}}\, \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{29}{4}}\, =\, \bigg(\, \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^a\, \bigg)^{-\frac{3}{4}} }\)

. . . . .\(\displaystyle \large{ \displaystyle \lim_{a \rightarrow \infty}\, \left(1\, +\, \dfrac{1}{a}\right)^{-\frac{29}{4}}\, =\, e^{-\frac{3}{4}}\, \cdot\, 1\, =\, e^{-\frac{3}{4}} }\)

That second line should be:

\(\displaystyle \displaystyle\lim_{a\to\ \infty}\bigg(1 + \frac{1}{a}\bigg)^{\tfrac{(-3a - 29)}{4}} \ \ = \ \ e^{\tfrac{-3}{4}}\cdot 1 \ \ = \ \ e^{\tfrac{-3}{4}} \)