the function f is given by ( f(x)=k + acdot b^{x} ), where ( alt0,bgt1 ), and k is a real number. which of…

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)