anonymous
  • anonymous
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..
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Zale101
anonymous
  • anonymous
I'm just confused whether it is 4;-2 or -2;4 @Zale101
anonymous
  • anonymous
What's the actual limit though ? I mean, find the limit as x approaches 2 bla bla bla - of what ?

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anonymous
  • 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..
anonymous
  • anonymous
|dw:1439684924826:dw|
anonymous
  • anonymous
anonymous
  • anonymous
That's what I am given @AngusV
anonymous
  • anonymous
@Zale101
anonymous
  • anonymous
Oh, well that's different.
anonymous
  • anonymous
So I figured out two limits are 4; -2 just not sure in what order.
anonymous
  • anonymous
Well, you know that line you see above and below the Ox axis is your function - the f of x - right ?
anonymous
  • anonymous
Yes
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
Ok...I understand how to find limits. Just not sure what order they go in after?
anonymous
  • 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
anonymous
  • anonymous
Oh ok!
anonymous
  • anonymous
Could you help me with a couple others? @AngusV
anonymous
  • anonymous
I'll do my best.
anonymous
  • anonymous
Ok thanks!
anonymous
  • anonymous
@AngusV ^
anonymous
  • anonymous
It is...whatever you can make out to be on the left side. Looks like 1.something.
anonymous
  • anonymous
1.3?
anonymous
  • 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?
anonymous
  • anonymous
@AngusV
anonymous
  • 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.
anonymous
  • anonymous
@AngusV
geerky42
  • 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.
anonymous
  • anonymous
Oh! So negative infinity; x=7?
anonymous
  • anonymous
Oh sorry, I meant negative infinity; x=9
geerky42
  • geerky42
Yep. You can imagine this way: \(\dfrac1{x-9}\) is same as \(\dfrac1x\), but shifted 9 units right.
geerky42
  • geerky42
So is everything clear?
anonymous
  • 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
anonymous
  • anonymous
@geerky42
geerky42
  • 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?
anonymous
  • 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?
geerky42
  • geerky42
Why negative infinity for right side?
anonymous
  • anonymous
I'm not sure. I'm confused by this entire question. @geerky42
geerky42
  • geerky42
Did you graph \(\dfrac1{(x-7)^2}\)?
anonymous
  • anonymous
Yes @geerky42
geerky42
  • geerky42
It looks something like this, right?|dw:1439687480641:dw| From left sides, what does it approaches to? Right sides?
geerky42
  • 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|

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