Continuity at a Point via Formulas - At A Glance

Answer

• When we try to evaluate f(-5), we find

  

  which is undefined. Since f(-5) does not exist, f is discontinuous at x = -5.

• When we try to evaluate f(-4), we find f(-4) = 0, which is fine. We'll see if  exists and agrees with f(-4). As x approaches -4 from the left, we use the part of the function definition that says The left-sided limit is

  When x approaches -4 from the right, x is just a bit bigger than -4, so we'll use the part of the function definition that says f(x) = x – 8 for -4 < x ≤ 2. Then the right-sided limit is

  

  Since the left- and right-hand limits agree,

  Alas, since f(-4) = 0, the limit disagrees with the function value, therefore f is discontinuous at x = -4.

•  f(0) = 0 – 8 = -8. As x approaches 0 from the left or the right, we use the part of the function definition that says f(x) = x – 8. Since the function is continuous at x = 0.
•  f(2) = 2 – 8 = -6, which is fine. Now we'll check the limit. As x approaches 2 from the left we use the part of the function's definition that says f(x) = x – 8 for -4 < x ≤ 2, and find   

  As x approaches 2 from the right we use the part of the function definition that says

  .

  The bottom of this expression factors as (x – 2)(x – 3), and from earlier work with limits we know that is undefined. Therefore  does not exist, and f is discontinuous at x = 2.

• When we try to evaluate f(3), we find which is undefined. Therefore f is discontinuous at x = 3.

Answer

• f is defined everywhere, so there can't be any values of x where f is discontinuous from being undefined. That means we need to look for places where the limit of f is undefined or disagrees with the value of f. The only place this could possibly happen is x = 0. Let's investigate. The left-hand limit is

The right-hand limit is

The one-sided limits agree,

Alas, this disagrees with the function value, f(0) = 0. The function is discontinuous at x = 0.

Answer

First, we'll look for values any of x where g is undefined. We're fine at x = 1, since g(1) = 2 by the second piece of the definition, but x = 2 is not fine. The function definition says what should happen for x < 2 and 2 < x, but doesn't make any provision for what happens at x = 2. Therefore g is undefined, and therefore discontinuous, at x = 2.

Now we will look for places where the limit is undefined or disagreeable. Since  is undefined, is also undefined and g is discontinuous when x = 1. That's the only troublesome spot we need to check.

To summarize, the function g is discontinuous at x = 1 and x = 2.

Answer

• The function definition says

The denominator factors, therefore we could rewrite this as

This function is undefined, and therefore discontinuous, at x = 4 and x = 5.The only place limits could go wrong, besides x = 4 and x = 5, is when x = 1. At x = 1 we have

As x approaches 1 from the right we have

Since the function isn't defined for values less than 1, we can't actually take a left-sided limit. In cases like this, we just use the right-sided limit as the value of the limit from all directions.

Therefore,

This agrees with h(1) = 1,

so the function is continuous at x = 1.

To summarize, h is only discontinuous at x = 4 and x = 5.