f(x) = (1 - x) / [(1 - x)(5 + x)]
My professor argues with us that the value 1 is considered a vertical asymptote.
Is he correct?
Sounds like it's time to stand up, show your evidence, and argue your point ... do you have face-to-face contact with this "professor", or is he/she some online hack? Don't be intimidated. What can he/she do to you? ... take away your birthday?
He is with me face to face in class. I feel that he made me wrong cuz he does not want to show students that he got a mistake, but if x = 1, will never be a vertical asymptote, i can discuss this issue with him again in his office. I just want to make sure he is wrong cuz if he is right, i am confused about this answer.
This is the problem we discussed in class.
Determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether f(x) goes to infinity or f(x) goes to negative infinity on either side of the asymptote.
f(x) = (1 - x) / (x^2 + x - 2)
fyi, this is not the same function you originally posted $\rightarrow$f(x) = (1 - x) / [(1 - x)(5 + x)]
$f(x) = \dfrac{1-x}{(x+2)(x-1)} = - \dfrac{x-1}{(x+2)(x-1)}$f(x) = (1 - x) / (x^2 + x - 2)
this function still has a point discontinuity at x = 1
May be this is the definition.
Examine lim 1/x as x goes to 0
Of course, we can draw a graph and compute a table of function values easily, by hand. While we say that the limits lim 1/x as x goes to 0+ and lim 1/x as x goes to 0- do not exist, they do so for different reasons. Specifically, as x goes 0+, 1/x increases without bound, while as x goes 0-, 1/x decreases without bound. To indicate this, we write
(1) lim 1/x as x goes 0+ = Infinity
and
(2) lim 1/x as x goes 0- = -Infinity
Graphically, this says that the graph of y = 1/x approaches the vertical line x = 0, as x goes to 0. When this occurs, we say that the line x = 0 is a vertical asymptote. It is important to note that while the limits (1) and (2) do not exist, we say that they equal Infinity and -Infinity, respectively, only to be specific as to why they do not exist. Finally, in view of the one-sided limits (1) and (2), we say as before that
lim 1/x as x goes 0 does not exist
$\displaystyle f(x) = \frac{{(1 - x)}}{{(1 - x)(5 + x)}} = \frac{1}{{(5 + x)}},\text{ if }x \ne 1$ then $\displaystyle \displaystyle{\lim _{x \to 1}}f(x) = \frac{1}{6}$
If your instructor sees this and then insists he/she is correct, then take the matter to the head of school.