1. ## what asymptote ?

Hello, I am having issues with an equation involving limits and I am suppose to find the Vertical Asymptote. I thought that this equation had no Vertical asymptotes but mathlab says there is a VA at x=4

f(x=) $\frac{x^2-9x+20}{x^2-8x+16}$

factor
$\frac{(x-5)(x-4)}{(x-4)(x-4)}$
the (x-4) cancels and leaves
$\frac{(x-5)}{(x-4)}$

which leave a possible assymptote at x = 4,
but if i plug in 4 to the original equation
$\frac{4^2-9(4)+20}{4^2-8(4)+16}$ = $\frac{0}{0}$
which is not really an assymptote, but a hole in the graph. What am I doing wrong?

2. ## Re: what asymptote ?

Hi,
i do see that there is vertical assymptote.please find below the reasoning
1> Vertical asymptotesThe line x = a is a vertical asymptote of the graph of the function y = ƒ(x) if at least one of the following statements is true:

a.lim x-> afor f(x) = + or - infinity
b.lim x-> -afor f(x) = + or - infinity.
2.keeping the first point in mind and applying L-Hospital's rule for f(x) it becomes (2x-9)/(2x-8).then put x=4 we get -1/0= -infinity
3.L-Hospital's rule;is a rule for converting an indeterminent form of functions (i.e 0/0,infinity/infinity,o^infinity,infinity^0,infinity-infinity......)into diterminent forms by differentiating the given function.