Determine if a Function is Continuous Using Limits - Pre-Calculus

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The gives us a hole at x=-3 so our function is not continuous at x=-3.

In order to determine if a function is continuous at a point three things must happen.

1. Taking the limit from the lefthand side of the function towards a specific point exists.

2. Taking the limit from the righthand side of the function towards a specific point exists.

3. The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

1. 

2. 

3. 

Because all of these conditions are met, the function is continuous at 0.

First, find that at any point where , .

Then find that

and .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

As approaches , the function approaches , which is undefined. However, if we factor , we get:

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from .

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The gives us a hole at x=-3 so our function is not continuous at x=-3.

In order to determine if a function is continuous at a point three things must happen.

1. Taking the limit from the lefthand side of the function towards a specific point exists.

2. Taking the limit from the righthand side of the function towards a specific point exists.

3. The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

1. 

2. 

3. 

Because all of these conditions are met, the function is continuous at 0.

First, find that at any point where , .

Then find that

and .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

As approaches , the function approaches , which is undefined. However, if we factor , we get:

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from .

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The gives us a hole at x=-3 so our function is not continuous at x=-3.

In order to determine if a function is continuous at a point three things must happen.

1. Taking the limit from the lefthand side of the function towards a specific point exists.

2. Taking the limit from the righthand side of the function towards a specific point exists.

3. The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

1. 

2. 

3. 

Because all of these conditions are met, the function is continuous at 0.

First, find that at any point where , .

Then find that

and .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

As approaches , the function approaches , which is undefined. However, if we factor , we get:

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from .

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The gives us a hole at x=-3 so our function is not continuous at x=-3.

In order to determine if a function is continuous at a point three things must happen.

1. Taking the limit from the lefthand side of the function towards a specific point exists.

2. Taking the limit from the righthand side of the function towards a specific point exists.

3. The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

1. 

2. 

3. 

Because all of these conditions are met, the function is continuous at 0.

First, find that at any point where , .

Then find that

and .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

As approaches , the function approaches , which is undefined. However, if we factor , we get:

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from .