The problem says:

If $$\lim_{x\to +\infty} \left\lbrack\frac{ax+1}{ax-1}\right\rbrack^x=9$$, determine $a$.

It appears to be a case of $\left\lbrack\frac{\infty}{\infty}\right\rbrack^\infty$. How can I solve such a case?


closed as off-topic by colormegone, user296602, choco_addicted, user91500, user21820 Apr 27 '16 at 8:08

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  • $\begingroup$ Do you know the laws of limit arithmetic? If $f(x)\to L$ then $g(f(x))\to ?$, assuming that both $g$ and $f$ are continuous? $\endgroup$ – user237392 Apr 27 '16 at 1:49
  • 1
    $\begingroup$ if $f(x)$ is your function, look at $\exp (\lim_{x\to\infty}\, \ln f(x))$. Is a case of $1^\infty$. $\endgroup$ – Chip Apr 27 '16 at 1:50

This is definetely a very interesting question. Here is what I think is a good way to think about it. First rearrange the expression (omitting limits for clarity):

$$\left( \frac{a x + 1 - 2 + 2 }{a x -1} \right)^x = \left( 1 + \frac{ 2 }{a x -1} \right)^x $$

Notice that, in the limit of $x \rightarrow ∞ $, the denominator reduces to $a x$. We rewrite the equation as:

$$\lim_{x\to \infty} \left( 1 + \frac{ 2 }{a x -1} \right)^x = 9 $$

or by noting what we have said about the denominator:

$$\lim_{x\to \infty} \left( 1 + \frac{ 2 }{a x} \right)^x = 9 $$

The LHS is exactly the definition of $e^{2/a}$, hence you can say that:

$$e^{2/a} = 9$$

from which you can easily find that:

$$a = \frac{2}{\ln{9}} = \frac{1}{\ln{(3)}}$$

I hope this helps you!

  • $\begingroup$ Oh, I was writing this as Dr. MV posted his. That is a good solution. $\endgroup$ – Stefano Apr 27 '16 at 2:22
  • $\begingroup$ yes absolutely right, I fixed it now. Thank you MathMajor. $\endgroup$ – Stefano Apr 27 '16 at 2:24
  • $\begingroup$ No problem. I like this approach. $\endgroup$ – MathMajor Apr 27 '16 at 2:30

Recall that the limit definition of the exponential function is

$$e^z=\lim_{n\to \infty}\left(1+\frac zn\right)^n \tag 1$$

Using $(1)$, we can write the limit of interest as

$$\begin{align} \lim_{x\to \infty}\left(\frac{ax+1}{ax-1}\right)^x&=\lim_{x\to \infty}\left(\frac{1+\frac{1/a}{x}}{1-\frac{1/a}{x}}\right)^x\\\\ &=e^{2/a} \end{align}$$

Finally, we have

$$e^{2/a}=9\implies a=\frac{1}{\log(3)}$$


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