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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?

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

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – colormegone, Community, choco_addicted, user91500, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
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    $\begingroup$ 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
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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!

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  • $\begingroup$ 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
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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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