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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..
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..

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just confused whether it is 4;2 or 2;4 @Zale101

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the actual limit though ? I mean, find the limit as x approaches 2 bla bla bla  of what ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439684924826:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I am given @AngusV

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, well that's different.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I figured out two limits are 4; 2 just not sure in what order.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, you know that line you see above and below the Ox axis is your function  the f of x  right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Put 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
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0Ok...I understand how to find limits. Just not sure what order they go in after?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Judging 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
 one year ago
Best ResponseYou've already chosen the best response.0Could you help me with a couple others? @AngusV

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is...whatever you can make out to be on the left side. Looks like 1.something.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use 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
 one year ago
Best ResponseYou've already chosen the best response.0And 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.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1For \(\displaystyle \lim_{x\to9^}\dfrac1{x9}\), 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh! So negative infinity; x=7?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh sorry, I meant negative infinity; x=9

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Yep. You can imagine this way: \(\dfrac1{x9}\) is same as \(\dfrac1x\), but shifted 9 units right.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1So is everything clear?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Not really. Problem is \(\large\displaystyle\lim_{x\to7}\dfrac1{(x7)^2}\) And you are basically saying that \(\large\displaystyle\lim_{x\to7}\dfrac1{(x7)^2}=\infty \) because \(\large\displaystyle\lim_{x\to7^}\dfrac1{(x7)^2}=\infty\) What about right side?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could 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
 one year ago
Best ResponseYou've already chosen the best response.1Why negative infinity for right side?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure. I'm confused by this entire question. @geerky42

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Did you graph \(\dfrac1{(x7)^2}\)?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1It looks something like this, right?dw:1439687480641:dw From left sides, what does it approaches to? Right sides?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Just 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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