Question

Determine whether the function is continuous at the given point \(c .\) If the function is not continuous, determine whether the discontinuity is removable or non-removable. $$ f(x)=\frac{4-x}{2-\sqrt{x}} ; c=4 $$

Short answer

The function is not continuous at \( c = 4 \); the discontinuity is removable.

Step-by-step solution

Identify the function and point

We are given the function \( f(x) = \frac{4-x}{2-\sqrt{x}} \) and need to check its continuity at the point \( c = 4 \). First, we'll examine whether the function is defined at \( x = 4 \).

Check if the function is defined at x = 4

Substitute \( x = 4 \) into the denominator of the function: \( 2 - \sqrt{4} = 2 - 2 = 0 \). Since the denominator becomes zero, the function is not defined at \( x = 4 \). Thus, initially, \( f(x) \) is not continuous at \( c = 4 \).

Investigate limits approaching c

Calculate the limit of \( f(x) \) as \( x \to 4 \). Change of variables helps: let \( h = \sqrt{x} - 2 \), then as \( x \to 4 \), \( h \to 0 \). So, \( x = (h + 2)^2 = h^2 + 4h + 4 \), substituting into \( f(x) \):\[\lim_{{h \to 0}} \frac{4 - (h^2 + 4h + 4)}{2 - (h+2)} = \lim_{{h \to 0}} \frac{-h^2 - 4h}{-h}\]

Simplify to find the limit

Simplify the expression:\[\lim_{{h \to 0}} \frac{-h^2 - 4h}{-h} = \lim_{{h \to 0}} h + 4 = 4\]Thus, the limit of \( f(x) \) as \( x \to 4 \) exists and is equal to 4.

Determine the type of discontinuity

Since the limit exists but \( f(4) \) is undefined, the discontinuity is removable. If the function can be redefined at \( x = 4 \) to be continuous, then this discontinuity is removable.

Key concepts

Limits and Continuity

When we talk about a function's continuity, we're asking if the function behaves 'nicely' without any unexpected jumps or breaks as we move along the curve. The concept of limits is essential in checking for this behavior.
The limit of a function at a particular point helps determine what value the function is approaching as we get closer to that point. In our exercise:

• We identify the point of interest, which is \( x = 4 \).
• Using limits, we analyze if the function approaches a finite value as \( x \to 4 \).

In the original solution, we found that \( \lim_{{x \to 4}} f(x) = 4 \).
Since the limit exists, this is a good start towards proving continuity. However, for the function to be fully continuous at \( x = 4 \), it must also be defined at this point, among other conditions.

Removable Discontinuity

A removable discontinuity occurs when the function's limit exists at a specific point, but the function itself is undefined at that point. It's like having a hole in the graph—if you imagine plotting it, there's a dot missing right at \( x = 4 \). For continuity:

• The function must be defined at the point.
• The limit as \( x \to 4 \) must equal the function's value at \( x = 4 \).

Our function \( f(x) = \frac{4-x}{2-\sqrt{x}} \) has been determined to have its denominator equal zero at \( x = 4 \), hence it is undefined there.
However, since the limit at \( x = 4 \) exists, we can "fill the hole" by redefining the function at this point. This makes the discontinuity removable, meaning it's a small fix, not a fundamental flaw in the function.

Discontinuous Points

Discontinuous points are locations on the graph of a function where there's a break or jump. These can occur for various reasons:

• The function's value is undefined.
• The left and right hand limits at the point are unequal.
• The limit does not exist.

In our problem, \( c = 4 \) is a discontinuous point because the function is undefined there due to the denominator being zero. This means that at \( x = 4 \), \( f(x) \) does not have a defined value. However, unlike other types of discontinuities like jump or infinite discontinuities, removable discontinuities are the simplest to resolve. It can be managed by changing the function so that it is defined appropriately at that point, thus "removing" the discontinuity.

Function Definition at a Point

For a function to be continuous at a point, it must be properly defined at that point first and foremost. Being 'defined' means that you can establish a solid value of the function at that given point \( x = c \).
Referring to our exercise, \( c = 4 \):

• The function at \( f(4) \) becomes undefined due to a zero in the denominator.
• This makes it impossible to directly calculate \( f(4) \).

To solve this issue, the function can be redefined. Since the limit as \( x \to 4 \) is verifiable and finite, we can redefine \( f(4) \) to be equal to the limit. Therefore, if we set \( f(4) = 4 \), then this resolves the issue of the discontinuity, making the function continuous at \( x = 4 \). Thus, defining the function correctly at a discontinuous point where the limit exists is vital to ensure continuity.