Does this function have vertical asymptotes? [ln(25-x^2]

Integrate

Integrate

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By my understanding this function very much has vertical asymptotes as ln can never equal zero.

So why does the solution manual say this function doesn't have vertical asymptote.

Vertical Asymptote.png

graph.png
 
Integrate said:
By my understanding this function very much has vertical asymptotes as ln can never equal zero.

So why does the solution manual say this function doesn't have vertical asymptote.

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View attachment 34730
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Clearly they are wrong. That happens.

The two limits are [imath]-\infty[/imath], not 0. (And they are one-sided limits.)

I don't know what they were thinking; those certainly are not holes!
 
Integrate said:
By my understanding this function very much has vertical asymptotes as ln can never equal zero.

So why does the solution manual say this function doesn't have vertical asymptote.

View attachment 34729

View attachment 34730
Click to expand...
From my understanding, holes are the removable discontinuity of the function.
For example,
[math]\dfrac{(2x-3)(x+1)(x-2)}{(x+2)(x+1)}[/math]
Since the numerator and denominator contain (x+1), we can cancel them out so we have a hole at (-1,15).
On the other hand, x=-2 is a vertical asymptote.

I think you're correct to call those vertical asymptotes instead of holes.
 
Integrate said:
By my understanding this function very much has vertical asymptotes as ln can never equal zero.

So why does the solution manual say this function doesn't have vertical asymptote.

View attachment 34729
Click to expand...
If you think that those limits equal 0, then what is the problem? 0 is a well defined value. The graphs (based on you thinking that the limits are 0) should NOT have a hole in it. The y-value, when x= +/- 5, is 0.

Of course, as Dr Peterson pointed out, those limits do not equal 0.
 
Dr.Peterson said:
Clearly they are wrong. That happens.

The two limits are [imath]-\infty[/imath], not 0. (And they are one-sided limits.)

I don't know what they were thinking; those certainly are not holes!
Click to expand...
Feels good to find errors. Like really good.
 
Beer induced ramblings follow.
Integrate said:
By my understanding this function very much has vertical asymptotes as ln can never equal zero.

So why does the solution manual say this function doesn't have vertical asymptote.
Click to expand...
Vertical Asymptote.png

Fortunately, the section for the Answers to All Odd-Numbered Exercises and Tests at the back of the book gives the correct answer (Review Exercise #75)

Screenshot_20230106-023421_Adobe Acrobat.jpg
 
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