To find the indeterminate form of \(\lim_{x \to \infty} \left(\frac{ax + 1}{ax - 1}\right)^x\), let's analyze the behavior of the function as \(x \to \infty\).
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### Step 1: Simplify the fraction
The given function is:
\[
f(x) = \left(\frac{ax + 1}{ax - 1}\right)^x
\]
Divide numerator and denominator inside the fraction by \(x\):
\[
f(x) = \left(\frac{a + \frac{1}{x}}{a - \frac{1}{x}}\right)^x
\]
As \(x \to \infty\), \(\frac{1}{x} \to 0\), so the fraction inside the parentheses approaches:
\[
\frac{a + 0}{a - 0} = \frac{a}{a} = 1
\]
Thus, the base approaches 1, and the exponent \(x \to \infty\). This gives the indeterminate form:
\[
1^\infty
\]
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### Step 2: Rewrite the expression using logarithms
Let:
\[
y = \left(\frac{ax + 1}{ax - 1}\right)^x
\]
Take the natural logarithm:
\[
\ln y = x \cdot \ln\left(\frac{ax + 1}{ax - 1}\right)
\]
### Step 3: Approximate the logarithmic term
For large \(x\), expand the fraction \(\frac{ax + 1}{ax - 1}\) using a binomial approximation:
\[
\frac{ax + 1}{ax - 1} = 1 + \frac{2}{ax - 1}
\]
For large \(x\), \(ax - 1 \approx ax\), so:
\[
\frac{ax + 1}{ax - 1} \approx 1 + \frac{2}{ax}
\]
Now substitute this back into \(\ln\):
\[
\ln\left(\frac{ax + 1}{ax - 1}\right) \approx \ln\left(1 + \frac{2}{ax}\right)
\]
Using the approximation \(\ln(1 + z) \approx z\) for small \(z\):
\[
\ln\left(1 + \frac{2}{ax}\right) \approx \frac{2}{ax}
\]
### Step 4: Substitute back and take the limit
Substitute this approximation into \(\ln y\):
\[
\ln y \approx x \cdot \frac{2}{ax} = \frac{2}{a}
\]
Exponentiate both sides:
\[
y = e^{\frac{2}{a}}
\]
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### Final Answer:
\[
\lim_{x \to \infty} \left(\frac{ax + 1}{ax - 1}\right)^x = e^{\frac{2}{a}}
\]
