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the function f is given by ( f(x)=k + acdot b^{x} ), where ( alt0,bgt1 …

Question

the function f is given by ( f(x)=k + acdot b^{x} ), where ( alt0,bgt1 ), and k is a real number. which of the following is true?
a ( lim_{x
ightarrow-infty}f(x)=-k )
b ( lim_{x
ightarrow-infty}f(x)=k )
c ( lim_{x
ightarrowinfty}f(x)=-k )
d ( lim_{x
ightarrowinfty}f(x)=k )

Explanation:

Step1: Analyze the limit as \(x\to-\infty\)

For \(y = b^{x}\) with \(b>1\), when \(x\to-\infty\), \(b^{x}=\frac{1}{b^{-x}}\to0\) (since \(b^{-x}\to+\infty\) as \(x\to-\infty\)).
So, \(\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(k + a\cdot b^{x})\).
Using the limit law \(\lim_{x\to c}(u(x)+v(x))=\lim_{x\to c}u(x)+\lim_{x\to c}v(x)\), we have \(\lim_{x\to-\infty}(k + a\cdot b^{x})=\lim_{x\to-\infty}k+a\lim_{x\to-\infty}b^{x}\).
Since \(\lim_{x\to-\infty}k = k\) (constant - limit) and \(a\lim_{x\to-\infty}b^{x}=a\times0 = 0\) (because \(b > 1\) and \(x\to-\infty\)), \(\lim_{x\to-\infty}f(x)=k\).

Step2: Analyze the limit as \(x\to+\infty\)

For \(y = b^{x}\) with \(b>1\), when \(x\to+\infty\), \(b^{x}\to+\infty\).
Since \(a<0\), \(a\cdot b^{x}\to-\infty\) as \(x\to+\infty\). Then \(\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}(k + a\cdot b^{x})\).
Using the limit law \(\lim_{x\to c}(u(x)+v(x))=\lim_{x\to c}u(x)+\lim_{x\to c}v(x)\), \(\lim_{x\to+\infty}(k + a\cdot b^{x})=\lim_{x\to+\infty}k+a\lim_{x\to+\infty}b^{x}\).
Since \(\lim_{x\to+\infty}k = k\) (constant - limit) and \(a\lim_{x\to+\infty}b^{x}=-\infty\) (because \(a<0\) and \(b^{x}\to+\infty\) as \(x\to+\infty\)), \(\lim_{x\to+\infty}f(x)=-\infty\).

Answer:

B. \(\lim_{x\to-\infty}f(x)=k\)