## anonymous one year ago 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

1. Luigi0210

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.

2. anonymous

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.

3. anonymous

@Luigi0210

4. Luigi0210

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. $$\large \lim_{x\rightarrow1.9} f(x)=100$$ $$\large \lim_{x\rightarrow1.99} f(x)=10000$$ $$\large \lim_{x\rightarrow1.999} f(x)=1000000$$ $$\large \lim_{x\rightarrow1.99999} f(x)=+\infty$$

5. anonymous

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?

6. Luigi0210

Are you asking how to clarify it is increasing?

7. anonymous

Yes. I

8. Luigi0210

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

9. Luigi0210

*f(1.9) not f(1)

10. anonymous

Refer to the attached plot.

11. anonymous

Thank you!