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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

@Zale101

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

|dw:1439684924826:dw|

- anonymous

##### 1 Attachment

- anonymous

That's what I am given @AngusV

- anonymous

@Zale101

- anonymous

Oh, well that's different.

- anonymous

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

- anonymous

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

- anonymous

Yes

- 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

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

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

- 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

Oh ok!

- anonymous

Could you help me with a couple others? @AngusV

- anonymous

I'll do my best.

- anonymous

Ok thanks!

##### 1 Attachment

- anonymous

@AngusV ^

- anonymous

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

- anonymous

1.3?

- 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

@AngusV

- 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

@AngusV

- 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

Oh! So negative infinity; x=7?

- anonymous

Oh sorry, I meant negative infinity; x=9

- geerky42

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

- geerky42

So is everything clear?

- 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

@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

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

Why negative infinity for right side?

- anonymous

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

- geerky42

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

- anonymous

Yes @geerky42

- geerky42

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

- 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|

Looking for something else?

Not the answer you are looking for? Search for more explanations.