Calc: Limits

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thecoolpharmboi

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10+ Year Member
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Hi,

So one question in the math section regarding limits: (I read the explanation, but pardon my stupidity, I still don't understand)

What is :
lim (x-> 1-) |x-1| / x-1 ?

The answer was -1. So... x-> 1- means x is approaching from below 1 right ? (as in x < 1)? The explanation of the question said that since x is less than 1,
x-1 must be negative. And so |x-1| = -(x-1) . But how can an absolute value of something be a negative?? If (x-1) is negative |x-1| should still be positive right??

T_T
I don't get it.
 
Hi,

So one question in the math section regarding limits: (I read the explanation, but pardon my stupidity, I still don't understand)

What is :
lim (x-> 1-) |x-1| / x-1 ?

The answer was -1. So... x-> 1- means x is approaching from below 1 right ? (as in x < 1)? The explanation of the question said that since x is less than 1,
x-1 must be negative. And so |x-1| = -(x-1) . But how can an absolute value of something be a negative?? If (x-1) is negative |x-1| should still be positive right??

T_T
I don't get it.

Basically just plug in -1 into the equation. So on the top -1-1 = -2... absolute value = 2. in the denominator -1-1 = -2. So 2/-2 = -1.
 
1- means approaching from less than 1. It makes sense because every number that you could plug in for x will equal a negative number with that equation. Its not saying an absolute value is negative, its saying x has to be negative. 2 different things, yo. Help any?
 
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It is. No matter what's in an absolute value, it's always positive. Dividing by a negative in the denominator is what gives you -1.

But that is not what they explained. They explained that :

|x-1| = -(x-1) thus:

lim (x->1-) |x-1| / x-1 ===> lim (x->1-) -(x-1) / x-1 ==>
- lim (x->1-) x-1 / x-1 => - lim (x->1-) 1 => -1 .

D: ????

If they said x-1 = negative...then I'd understand. But for some reason that is not what they said. They said to examine " |x-1| " first.
 
if something is l???l then if the number inside the l???l is negative then itll become -(???) it is positive its just (???)

in your case you got lx-1l / (x-1) with it approaching the lower end of 1

this means that it approaches one but doesnt equal one so just take the number as 0.99

if u plug it in it becomes (0.99-1) = -0.01

so since its a negative the lx-1l becomes -(x-1)

when you deal with limits the negative goes outside the limit so it become

-lim (x-1)/(x-1) then plug in 0.99 or whatever and since they are the same it equals 1 and just plug the negative back in

hope that helps
 
if something is l???l then if the number inside the l???l is negative then itll become -(???) it is positive its just (???)

in your case you got lx-1l / (x-1) with it approaching the lower end of 1

this means that it approaches one but doesnt equal one so just take the number as 0.99

if u plug it in it becomes (0.99-1) = -0.01

so since its a negative the lx-1l becomes -(x-1)

when you deal with limits the negative goes outside the limit so it become

-lim (x-1)/(x-1) then plug in 0.99 or whatever and since they are the same it equals 1 and just plug the negative back in

hope that helps

So...if the number is -1, for example. If |1| = 1, then |-1| = -1 ?!?!?!

@_@ since u said if the number in the |???| is negative, then it becomes -(???) right? But I thought |???| is absolute value of something, so it's always positive...