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

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?

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

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

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:

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

• Sorry, i didn't understand what you did – Mycroft Jul 5 at 4:20
• @Mycroft The same idea as this post. – Bach Jul 5 at 4:24

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

• Could you add some steps between the standard form and L=e^(−2/a)? – Mycroft Jul 5 at 4:23

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 $$a=\frac1{\log(1/3)}$$.

• You have missed x below the t – Archis Welankar Jul 5 at 4:35