the lim [as x goes to infinity] (.25)^x =0??

Yes, the limit is zero. The more times you multiply a number by 0.25 (which is what is happening when x goes to infinity), the smaller is the result.

To formally prove that the limit as x approaches infinity of (0.25)^x is indeed zero, we can use the definition of a limit.

Let's denote the function as f(x) = (0.25)^x. To find the limit, we need to evaluate what happens to f(x) as x approaches infinity.

As x gets larger and larger, the value of f(x) will get smaller and smaller because we are continuously multiplying it by a number less than 1 (0.25). The behavior of f(x) as x approaches infinity can be best understood by rewriting (0.25)^x as 1/(4^x), which has the same value.

Now, let's calculate the limit:

lim(x → ∞) (0.25)^x = lim(x → ∞) (1/(4^x)).

To evaluate this limit, we can rewrite it using properties of limits:

lim(x → ∞) (1/(4^x)) = 1 / lim(x → ∞) (4^x).

The denominator (4^x) represents an exponential function that grows without bound as x goes to infinity. Therefore, the limit of 4^x as x approaches infinity is infinity:

lim(x → ∞) (4^x) = ∞.

Taking the reciprocal of this infinity gives us:

1 / ∞ = 0.

So, the limit as x approaches infinity of (0.25)^x is indeed equal to zero:

lim(x → ∞) (0.25)^x = 0.