Question

Please help one question!

Use graphs and tables to find the limit and identify any vertical asymptotes of the function:

lim x->2, 1/((x-2)^2)

I know that it is +infinite and the asymptote is 2 but i dont know how to explain it

Answer 1

I guess you could say that as you approach 2 from the left it shoots up to infinity. And being 2 would leave you with an undefined value. The limit coming from the right to 2 would also approach infinity.

Answer 2

But i dont know how to explain why it does that. Like I need to explain how I got my answer but I dont know how to explain it.

Answer 3

@Luigi0210

Answer 4

You could do what the instructions said and use a graph and table. Visually speaking, you can see it starts shooting up as it approaches 2.

Mathematically, if you look at a table, the values start increasing as well.
\(\large \lim_{x\rightarrow1.9} f(x)=100\)

\(\large \lim_{x\rightarrow1.99} f(x)=10000\)

\(\large \lim_{x\rightarrow1.999} f(x)=1000000\)

\(\large \lim_{x\rightarrow1.99999} f(x)=+\infty\)

Answer 5

I understand that part thank you, but what part of the equation shows the graph going up? I know that x -> 2 shows us the vertical asymptote of 2, but how does 1/((x-2)^2) show it going up?

Answer 6

Are you asking how to clarify it is increasing?

Answer 7

Yes. I

Answer 8

plug in values?

x=1, \(f(x)=\frac{1}{(x-2)^2} ===> f(1)\frac{1}{(1-2)^2}=\frac{1}{-1^2}=1\)

x=1.5 \(f(x)=\frac{1}{(x-2)^2} ===> f(1.5)\frac{1}{(1.5-2)^2}=\frac{1}{-.5^2}=4\)

x=1.75 \(f(x)=\frac{1}{(x-2)^2} ===> f(1.75)\frac{1}{(1.75-2)^2}=\frac{1}{-.25^2}=16\)

x=1.9 \(f(x)=\frac{1}{(x-2)^2} ===> f(1)\frac{1.9}{(1.9-2)^2}=\frac{1}{-.1^2}=100\)

Answer 9

*f(1.9) not f(1)

Answer 10

Refer to the attached plot.

Answer 11

Thank you!