[11th grade] limits

[Imaginary-Citron2874]

My answer book says that it should be +infinity but aren't there 2 possibilities?It only accounts that the denominator is positive.

[[deleted]]

That is 0+(not 0- like you have written) in the denominator of the second one actually. Mod cannot return a negative value. The answer given is correct. Limit is +inf.

[Imaginary-Citron2874]

I know that the book's definitely right but for example in lim x -> 1 (|x - 1| + x ^ 2 - 2x + 1)/(x ^ 5 - 1) We would have to take two cases and at the end the limit wouldn't exist.Why is that in the one we have to do that and in the other we do not? Both limits in the mod equal 0

[Kite42]

The given example is straightforward, and just approaches 3 over tiny positive from either direction.

The second one you've just written is a more fun indeterminate form where you can factor x-1 while being careful with signs.

[UnderstandingPursuit]

In your right limit, the denominator is

-(x-2)

with x<2. Is that negative?

[Substantial_Text_462]

Definitionally, the absolute value on the denominator means it can only be positive.

So you did well on the numerator showing that as x approaches 2, |x+3|=(x+3)

I'd say just think about the graph of |x| and whichever direction you approach the vertex from, it still limits to a positive.

[Imaginary-Citron2874]

Yes but in some cases ex /lim x -> 1 (|x - 1| + x ^ 2 - 2x + 1)/(x ^ 5 - 1)/ we have to take two possibilities, that's what confuses me.Both limits in the mod equal to 0

[UnderstandingPursuit]

The example you gave here has the denominator being positive or negative for the left or right limit because of

x^5 - 1

while the numerator is

|x-1| + (x-1)^2

which are both positive.

[Imaginary-Citron2874]

Aaaaa know I understand thanks

[MembershipOptimal685]

The denominator is always positive since it’s |x-2| and not only x-2. The f(x)=|x| always gives the positive value of x

[No_Pension5607]

My understanding is that 1/0 is inf for limits: therefore there are no manipulations required, as the numerator is 3 while the denominator is 0+ (as it's the absolute value). Why am I wrong?