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What is the value of $a$ in this limit? $$ \lim_{x\to\infty}\left(\frac{ax-1}{ax+1}\right)^x=9 $$

I'm preparing for an exam and found this question on a list, but I don't even know how to start.

Could someone help me?

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

So if this equals $9$, then $e^{-2/a} = 9$ or $e^{-1/a} = 3$ so $-1/a = \ln(3) $ or $a = \dfrac{-1}{\ln(3)} $.

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Hint:

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

Further hint:

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

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    $\begingroup$ Sorry, i didn't understand what you did $\endgroup$ – Mycroft Jul 5 at 4:20
  • $\begingroup$ @Mycroft The same idea as this post. $\endgroup$ – Bach Jul 5 at 4:24
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This is a standard limit of the form $1^\infty$.

It is evaluated as $$L=\lim_{x \to \infty}e^{(f(x)-1)(g(x)}$$ where $f(x)$ is the numerator function and $g(x)$ is the denominator function. You can prove this by taking log of the limit.

So applying it gives the answer as $$L=e^{\frac{-2}{a}}$$. Since $L=9$ so $$a=\frac{-2}{ln9}$$.This question is helpful

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  • $\begingroup$ Could you add some steps between the standard form and L=e^(−2/a)? $\endgroup$ – Mycroft Jul 5 at 4:23
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Note that $$\lim_{x\to\infty}\left(1+\frac{t}{x}\right)^x=e^t$$ Hence, we have

$$\begin{align} \lim_{x\to\infty}\left(\frac{ax-1}{ax+1}\right)^x&=\lim_{x\to\infty}\left(\frac{1+\frac{-1/a}{x}}{1+\frac{1/a}{x}}\right)^x\tag1 &=e^{-2/a} \end{align}$$

Now, set the right-hand side of $(1)$ to $9$ and solve for $a$ to find $a=\frac1{\log(1/3)}$.

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    $\begingroup$ You have missed x below the t $\endgroup$ – Archis Welankar Jul 5 at 4:35

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