HELPPP - limits

paofu

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Oct 16, 2020
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Why is the answer equal to 1?????????
I am so confused

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Thank you😿😿😿
 
Last edited:

Otis

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Apr 22, 2015
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Hi paofu. Your work looks good, except there's a mistake in the first one-sided limit. In each limit, x is approaching -2, so we consider only values of x that are very close to -2. That means x is always less than 0, whether we approach from the left or the right. Therefore, we replace |x| with -x, and the one-sided limits are both the same.

By the way, we cannot cancel x on top and bottom, like you did in the last line. (We cancel only factors, not numbers being added or subtracted.)

We don't need a cancellation to see that (2 + x)/(2 + x) is always 1 (because the numerator and denominator are the same). We never get 0/0 because -2 is not in the function's domain.

😎
 

HallsofIvy

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Your equation, \(\displaystyle \frac{2- |x|}{2+ x}=\frac{2- x}{2+x}\) is valid for x positive. But if x is close to -2 it is NOT positive! Taking the limit "as x goes to -2", we are only interested in x close to, but not equal to, -2. What is true for x positive is irrelevant!
 

pka

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paofu

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Oct 16, 2020
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Hi paofu. Your work looks good, except there's a mistake in the first one-sided limit. In each limit, x is approaching -2, so we consider only values of x that are very close to -2. That means x is always less than 0, whether we approach from the left or the right. Therefore, we replace |x| with -x, and the one-sided limits are both the same.

By the way, we cannot cancel x on top and bottom, like you did in the last line. (We cancel only factors, not numbers being added or subtracted.)

We don't need a cancellation to see that (2 + x)/(2 + x) is always 1 (because the numerator and denominator are the same). We never get 0/0 because -2 is not in the function's domain.

😎
I get it! Thank you so much for your help 😊 !!!
I didn’t cancel the x, but (x+1), I guess it’s ok then:)

Have a nice day!!
 

paofu

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Oct 16, 2020
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Your equation, \(\displaystyle \frac{2- |x|}{2+ x}=\frac{2- x}{2+x}\) is valid for x positive. But if x is close to -2 it is NOT positive! Taking the limit "as x goes to -2", we are only interested in x close to, but not equal to, -2. What is true for x positive is irrelevant!
Got it! Thank you!
 

Otis

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Apr 22, 2015
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2,562
… I didn’t cancel the x, but (x+1), I guess it’s ok then …
Hi. I'm not sure what you did because all I see is the line drawn through x. Did you express 2+x as (x+1)+1 in your head and then cancel the (x+1) parts? That would not be a valid cancellation, either.

As I'd mentioned, a cancellation isn't needed.

😎
 
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