# Finding a constant that satisfies a limit

• Nov 8th 2010, 09:55 AM
KelvinScc
Finding a constant that satisfies a limit
Hi, I was given the following question, and I am not entirely sure how to approach it.

Find the constant a such that

lim x->infinity ((x+a)/(x-a))^x=e

I have tried applying l'hopitals rule, and substituting x=1/y so that the limit becomes

lim y->0 ((1+ay)/1-ay))

However, I keep getting stuck. Guidance would be much appreciated.

Thank you very much!
• Nov 8th 2010, 10:24 AM
Chris L T521
Quote:

Originally Posted by KelvinScc
Hi, I was given the following question, and I am not entirely sure how to approach it.

Find the constant a such that

lim x->infinity ((x+a)/(x-a))^x=e

I have tried applying l'hopitals rule, and substituting x=1/y so that the limit becomes

lim y->0 ((1+ay)/1-ay))

However, I keep getting stuck. Guidance would be much appreciated.

Thank you very much!

Observe that $\displaystyle \dfrac{x+a}{x-a}=\dfrac{x-a+2a}{x-a}=1+\dfrac{2a}{x-a}$

So now keep in mind the fact that $\displaystyle \lim\limits_{u\to\infty}\left(1+\dfrac{1}{u}\right )^u=e$.

Messing around with the original limit, we get

\displaystyle \begin{aligned}\lim\limits_{x\to\infty}\left(1+\df rac{1}{(x-a)/(2a)}\right)^x &=\lim\limits_{x\to\infty}\left(1+\dfrac{1}{(x-a)/(2a)}\right)^{2a[(x-a)/(2a)]+a}\\ &=\left[\lim\limits_{x\to\infty}\left(1+\dfrac{1}{(x-a)/(2a)}\right)^{(x-a)/(2a)}\right]^{2a}\cdot\lim\limits_{x\to\infty}\left(1+\dfrac{1 }{(x-a)/(2a)}\right)^a\\ &= e^{2a}\cdot 1\end{aligned}

Thus, for $\displaystyle \lim\limits_{x\to\infty}\left(\dfrac{x+a}{x-a}\right)^x=e$, we must have $\displaystyle e^{2a}=e\implies a=\frac{1}{2}$.

Does this make sense?
• Nov 8th 2010, 12:34 PM
KelvinScc
Thanks!~
Yes, it makes complete sense. Thank you so much(Clapping)