finding a limit function that gives a vertical asym.and limi

chandler

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Dec 3, 2008
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I need a little help. this project is due in a day and I have been playing around with numbers all thanksgiving break to figure this out..

the problem is i need a function that gives a vertical asymp. of 8 with a limit of 2. there are no directions except to find the funtion and sketch the graph..

so far the only thing i could come up with is random numerator numbers and a denominator of (x-8)
 
"limit of 2" is vague ... do you mean that the function has a limit equal to 2 as x approaches a specific fixed value, as x approaches infinity, ... what???

please state the problem exactly as it was given to you.
 
the problem is find a function that has a vertical asymtope of 8 with the limit as x aprroaches 2
 
chandler said:
the problem is find a function that has a vertical asymtope of 8 with the limit as x aprroaches 2

if this is how the problem was stated, then it still does not make sense.

first of all, a vertical asymptote is a line of the form x = k ... so, is your vertical asymptote x = 8 ?

second, what is the limit of f(x) as x approaches 2 ?

in other words, \(\displaystyle \lim_{x \to 2} f(x) = ?\)
 
yes i know it sounds confusing but that is what my teacher has told me to find. an yes you are correct on the limit.
 
I am guessing here, but I think the teacher may have meant find a function with a vertical asymptote at x=8 and a horizontal asymptote at y=2.

Just a thought. Try that.
 
Can you post the exact problem that your teacher handed you - verbatim.

Or did he assign the problem verbally?
 
the teacher gave a paper but the directions say determine the function that has a vertical asymptope of 8 and then find the limit. but the limit has to be to because we had to use our lunch codes. mine is 93528 the first letter which is the vert. asymp. is the most non-zero number to the right and the limit is the most non-zero number to the left which is 2. and no there can be no horizontal asymp.. but i had to turn the project in today. thank you all for your time and effort. i apologize, that i didnt state the question right the first time.
 
I'll be interested in looking at the solution when it is provided to you.
 
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