# prove f(x)=b^x is continuous on (-infinity,infinity) i.s at all a exist in reals. here assume b does not equal 1, but b>0

## To prove that the function f(x) = b^x is continuous on (-∞, ∞) for all real numbers a, we need to show that it satisfies the definition of continuity.

Recall that a function f(x) is continuous at a point c if three conditions are met:

1. f(c) is defined,

2. The limit of f(x) as x approaches c exists, and

3. The limit of f(x) as x approaches c is equal to f(c).

Now let's go through each condition step by step:

1. f(c) is defined:

For any real number c, f(c) = b^c is defined because b^c is defined for all real numbers.

2. The limit of f(x) as x approaches c exists:

To prove that the limit exists, we need to show that the left-hand limit and the right-hand limit both exist and are equal.

Let's first consider the left-hand limit. As x approaches c from the left side (x < c), the function b^x approaches b^c. This can be shown by taking the natural logarithm of both sides of the equation: ln(b^x) = ln(b^c), which simplifies to x * ln(b) = c * ln(b). As x approaches c, ln(b) is a constant, so x approaches c * ln(b)/ln(b), which is just c. Therefore, the left-hand limit is b^c.

Next, let's consider the right-hand limit. As x approaches c from the right side (x > c), the function b^x also approaches b^c. This can be shown using a similar argument as above.

Since the left-hand limit and the right-hand limit both exist and are equal to b^c, the limit of f(x) as x approaches c exists.

3. The limit of f(x) as x approaches c is equal to f(c):

We've already established that the limit of f(x) as x approaches c is b^c. And we know that f(c) = b^c. Therefore, the limit of f(x) as x approaches c is equal to f(c).

By satisfying all three conditions, we have proved that the function f(x) = b^x is continuous on (-∞, ∞) for all real numbers a.