## anonymous one year ago Need help on a couple Limit Problems, precalculus. Find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x..

1. anonymous

@Zale101

2. anonymous

I'm just confused whether it is 4;-2 or -2;4 @Zale101

3. anonymous

What's the actual limit though ? I mean, find the limit as x approaches 2 bla bla bla - of what ?

4. anonymous

Find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x..

5. anonymous

|dw:1439684924826:dw|

6. anonymous

7. anonymous

That's what I am given @AngusV

8. anonymous

@Zale101

9. anonymous

Oh, well that's different.

10. anonymous

So I figured out two limits are 4; -2 just not sure in what order.

11. anonymous

Well, you know that line you see above and below the Ox axis is your function - the f of x - right ?

12. anonymous

Yes

13. anonymous

Put yourself "in the position" of that line as you are approaching x=2 from below: The simplest way to imagine this would be to consider "yourself" as 1.99999999.... Since you're less than 2 now, you would have to find yourself on the above line, where f(x)=4. Thus, the limit of f(x) when you are approaching 2 from "below" is 4.

14. anonymous

If you were to approach x=2 from "above" - imagine yourself as being 2.00001 and you see that for x=2.000001, f of x or y=-2.

15. anonymous

Ok...I understand how to find limits. Just not sure what order they go in after?

16. anonymous

Judging by the phrasing, it says : find the limit bla bla bla as x approaches from the left, find the limit bla bla bla as x approaches from the right. So since the answer for the left approach is 4 and for the right approach is -2 It would be 4;-2

17. anonymous

Oh ok!

18. anonymous

Could you help me with a couple others? @AngusV

19. anonymous

I'll do my best.

20. anonymous

Ok thanks!

21. anonymous

@AngusV ^

22. anonymous

It is...whatever you can make out to be on the left side. Looks like 1.something.

23. anonymous

1.3?

24. anonymous

Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 9 as x approaches 9 from the left. Would it be positive infinity; x=9?

25. anonymous

@AngusV

26. anonymous

And for: Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 7 squared as x approaches 7. Would this be ok? Using a graph, I see that as you approach 7 from the left, then you get positive infinity.

27. anonymous

@AngusV

28. geerky42

For $$\displaystyle \lim_{x\to9^-}\dfrac1{x-9}$$, no it's not positive infinity. You are correct about 7 part, but problem just said "x approaches" 7, which means x approaches 7 from BOTH sides.

29. anonymous

Oh! So negative infinity; x=7?

30. anonymous

Oh sorry, I meant negative infinity; x=9

31. geerky42

Yep. You can imagine this way: $$\dfrac1{x-9}$$ is same as $$\dfrac1x$$, but shifted 9 units right.

32. geerky42

So is everything clear?

33. anonymous

Yes, just that other problem. Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 7 squared as x approaches 7. Would this be ok? Using a graph, I see that as you approach 7 from the left, then you get positive infinity. Just wanted to make sure I worded it correctly

34. anonymous

@geerky42

35. geerky42

Not really. Problem is $$\large\displaystyle\lim_{x\to7}\dfrac1{(x-7)^2}$$ And you are basically saying that $$\large\displaystyle\lim_{x\to7}\dfrac1{(x-7)^2}=\infty$$ because $$\large\displaystyle\lim_{x\to7^-}\dfrac1{(x-7)^2}=\infty$$ What about right side?

36. anonymous

Could I say: Using a graph, I see that as you approach 7 from the left, then you get positive infinity. When you approach 7 from the right side, then you get negative infinity?

37. geerky42

Why negative infinity for right side?

38. anonymous

I'm not sure. I'm confused by this entire question. @geerky42

39. geerky42

Did you graph $$\dfrac1{(x-7)^2}$$?

40. anonymous

Yes @geerky42

41. geerky42

It looks something like this, right?|dw:1439687480641:dw| From left sides, what does it approaches to? Right sides?

42. geerky42

Just imagine dot approaching to x=7 (on left side for now), what value of y would this dot approaches to?|dw:1439687675392:dw|