13. Which function will not approach 0 as a increases? a. $$ f(x)=3 \cdot\left(\frac{1}{8}\right)^{3} $$ b. $$ f(x)=5 \cdot\left(\frac{s}{4}\right)^{3} $$ c. $$ f(x)=2 \cdot\left(\frac{1}{2}\right)^{3} $$ d. $$ f(x)=4 \cdot 5^{x} $$

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To determine which function does not approach 0 as its variable increases (that is, as x → ∞), we need to look at the base of the exponential part in each function.

a. f(x) = 3 · (1/8)³  
  This expression is a constant value because the term (1/8)³ is computed first and is not a variable exponent. Regardless, if the exponent were a variable and 1/8 < 1, it would decay toward 0. Here, though, the function is simply constant.

b. f(x) = 5 · (s/4)³  
  This also appears to be a constant. Assuming "s" is a constant, (s/4)³ is just a constant value, and 5 times that constant remains constant. (If s were meant to be a variable, it is written differently from the others. However, typically, a variable in an exponential decay context would be in the exponent.)

c. f(x) = 2 · (1/2)³  
  This is similarly a constant value. 1/2 < 1 implies that if the exponent were a variable, the function would decay to 0. However, here the exponent is 3, so the function does not change—it is just a fixed number.

d. f(x) = 4 · 5ˣ  
  Here, the variable x is in the exponent. Since the base 5 is greater than 1, as x increases the function value grows exponentially rather than approaching 0.

Thus, when considering the behavior as x increases, it is option d, f(x) = 4 · 5ˣ, that does not approach 0 but instead increases without bound.
Option d, $$ f(x) = 4 \cdot 5^{x} $$, does not approach 0 as x increases.

13. Which function will not approach 0 as a increases? a. \( f(x)=3 \cdot\left(\frac{1}{8}\right)^{3} \) b. \( f(x)=5 \cdot\left(\frac{s}{4}\right)^{3} \) c. \( f(x)=2 \cdot\left(\frac{1}{2}\right)^{3} \) d. \( f(x)=4 \cdot 5^{x} \)

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