Question

the exponential function ( f(x)=left(\frac{1}{6}
ight)^{x}+9 ) has the horizontal asymptote ( y = ). this means that as ( x \to infty ), we have ( left(\frac{1}{6}
ight)^{x}+9 \to )

Explanation:

Step1: Analyze the limit of the exponential term

For the exponential function \(y = a^{x}\), when \(0 < a<1\) and \(x\to\infty\), \(\lim_{x
ightarrow\infty}a^{x}=0\). Here \(a = \frac{1}{6}\), so \(\lim_{x
ightarrow\infty}(\frac{1}{6})^{x}=0\).

Step2: Find the limit of the function \(f(x)\)

We know that \(f(x)=(\frac{1}{6})^{x}+9\). By the limit rule \(\lim_{x
ightarrow\infty}(u(x)+v(x))=\lim_{x
ightarrow\infty}u(x)+\lim_{x
ightarrow\infty}v(x)\). Let \(u(x)=(\frac{1}{6})^{x}\) and \(v(x) = 9\). Then \(\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(\frac{1}{6})^{x}+\lim_{x
ightarrow\infty}9\). Since \(\lim_{x
ightarrow\infty}(\frac{1}{6})^{x}=0\) and \(\lim_{x
ightarrow\infty}9 = 9\), we have \(\lim_{x
ightarrow\infty}f(x)=9\).

Answer:

\(9\)