Jason asked in Science & MathematicsMathematics · 1 month ago

# What will happen if the Removable Discontinuity is the same as the Vertical Asymptote?

Example: f(x)=16-4x^2/x^2-4x+4

It is both x=2

Update:

Example: f(x)=(16-4x^2)/(x^2-4x+4)

It is both x=2

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• Vaman
Lv 7
1 month ago

f(x)=(16-4x^2)/(x^2-4x+4)= (4-2x)(4+2x)/(x-2)^2=4(2-x)(2+x)/(2-x)^2= 4(2+x)/(2-x)

x=2 is the singular point which can not be removed.

• 1 month ago

f(x)=(16-4x^2)/(x^2-4x+4)

=>

f(x)=4(2+x)(2-x)/(2-x)^2

=>

f(x)=4(2+x)/(2-x), if x=/=2.

So, if the vertical asymptote x=2 exists, the

discontinuity at x=2 exists too & vice versa.

It is obvious that if the discontinuity is removed--

that means the curve is continuous at that point.

Thus, no vertical asymptote exists there.

• 1 month ago

Hello I recently answered this problem in a test, however I only got 5 out of 10 points because my teacher said that there was a removable discontinuity even though I think that only a infinite discontinuity exist... I just want to know what is correct

• 1 month ago

It is not acceptable for someone studying calculus to have trouble with formatting fractions. Do try to focus.

Generally, a removable discontinuity is called removable because, you can make the function continuous by adding just one point (it looks like a hole).

So a vertical asymptote is not a removable discontinuity.

(16-4x^2) / (x^2-4x+4) = (4+2x) / (x-2) goes to infinity as x->2 so it's a vertical asymptote.

The definition may differ from textbook to textbook though.

• 1 month ago

A simpler example would be the function f(x) = x/x

I'd say it's undefined at x=0, but in the limit is equal to 1 at that point.

• 1 month ago

Except for x=0, you have given us the equation of a line.

Let's simplify the function you provided. Division takes precedence.

16 - (4x²/x²) - 4x + 4

16 - 4 - 4x + 4

-4x + 16

There's a hole at x=0, but no asymptote.

Perhaps you meant (16 - 4x²)/(x² - 4x + 4)?

-4(x - 2)(x + 2)/[(x - 2)²]

= -4(x + 2)/(x - 2)

x = 2 is not in the domain, but that's okay because you still have another x - 2 in the denominator to create a vertical asymptote at x=2.

(You also have a horizontal asymptote at y=-4)

• rotchm
Lv 7
1 month ago

It typically makes a hole. But not always... that's because...

well, you need to learn to write correctly. Did you really mean

f(x)=16-4x^2/x^2-4x+4 ?  (no you didn't btw).

Correct that and we will explain further.

To you own reply. No, there are no removable discontinuity in your function.