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Mathematics 7 Online
TheConMan:

Math help plz

Ballery1:

|dw:1568431434243:dw|

JSVSL7:

So first you have to find the left and right hand limits as x approaches a, given as 1. So, from the left, the values are LESS than 1. So the equation that corresponds to the piecewise LESS than 1 is the first one. From the right, the values are GREATER than 1, so the equation should be GREATER than 1 or the second one. Get what I mean?

Ballery1:

|dw:1568431569935:dw|

JSVSL7:

No, you plug in 1 simply to the corresponding equation

Ballery1:

i was gonna draw the piecewise funcs

JSVSL7:

Ah sure

Ballery1:

|dw:1568431752013:dw|

JSVSL7:

This is what the graph should look like https://www.desmos.com/calculator/vji9v6sfso Remember the open and closed circles

JSVSL7:

You're a bit off and there is no open circle for x^2 to indicate >

Ballery1:

icy

Ballery1:

|dw:1568432152594:dw|

JSVSL7:

Why did you plug in 2 for x^2 ?

Ballery1:

they both approach 2 different limits

Ballery1:

yes, since it said x>1 so i started with 2

Ballery1:

2^2, 3^2, 4^2 and so on

JSVSL7:

No. Recall that limits work differently from unincluded domains. Limits still consider them. That's why they're called the limit. You would plug in 1, since it's asking you to find the limit of x approaching 1.

JSVSL7:

They do have different limits and thus the limit of x approaching 1 is DNE, but your right hand limit of 4 is incorrect.

Ballery1:

how? it said x>1...oh nvm /.-

Ballery1:

|dw:1568432474375:dw|

Ballery1:

k thank you sir. i really appreciated your help

JSVSL7:

np

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