Can the vertical asymptote be the same as the hole?

Question

amistre64 · Answer

similar, but no

amistre64 · Answer

an asymptote is something that you slide up next to ..... a hole is something you can jump over

anonymous · Answer

It you look at the function $$
\frac{1}{x^2}
$$You can see that both limits converge at $\infty$
You can't really fill in such a "hole" because $\infty$ is not a number.

amistre64 · Answer

$$f(x)=\frac{(x-a)}{(x-a)(x-b)}$$
a hole can be thought of as having a control in the bottom of the fraction .... hole values can cancel each other out until they reach zero, then its undefined

an asymptote has no control .... and just slides into infinity

directrix · Answer

@helloihaveaquestion 

Graph: y = 4/x^2.
Graph: y = (4x)/x.

One has a vertical asymptote.
The other has a hole.

anonymous · Answer

which is which?

directrix · Answer

@helloihaveaquestion 

That is the question I had in mind that you would answer so that you could see what caused a hole and what caused a vertical asymptote.

anonymous · Answer

well 4x/x cant have a VA because there is a zero in the denominator which is the same zero as one in the numerator. but its a straight line...

directrix · Answer

y = 4x/x is not the same as the line y = 4.  Draw the graphs.

anonymous · Answer

dont the x's cross out

directrix · Answer

Yes. If x = 0, then you would be dividing by 0 which is not allowed.

Y = 4x/x is the same as y = 4, where x is not 0. I'll draw. the graph.

directrix · Answer

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