Find the limit, lim x--3, x^2 + 2x -3 / x^2 - x
9 - 6 - 3 / 9 - 3
= 0/6
= 0
this is indeterminate ..
lim x -- 3 , (x+3)(x-1) / x(x-1) =
x+3/x =
= 6/3
= 2

Question

Find the limit, lim x-->3, x^2 + 2x -3 / x^2 - x
9 - 6 - 3 / 9 - 3
= 0/6
=  0 
this is  indeterminate ..
lim x --> 3 , (x+3)(x-1) / x(x-1) = 
x+3/x = 
 = 6/3 
= 2

shamil98 · Answer

$$\lim_{x \rightarrow 3} \frac{ x^2 +2x -3 }{ x^2 - x }$$

nincompoop · Answer

there you go

nincompoop · Answer

the first rule is to plug and chug. go ahead and plug 3 into all of your x's and solve
the only restriction you will have is if your denominator becomes zero. if that were the case, then you will ditch the plug and chug because that is not the solution.

anonymous · Answer

^^

shamil98 · Answer

Alright, thanks.

nincompoop · Answer

from the looks of it, your denominator is not zero. so you have a well-defined function that doesn't pose too much of a problem to work with.

kira_yamato · Answer

Indeterminate forms would be things like 0^0, 0/0, ∞/∞ etc. In these cases, use L'Hopital's Rule

anonymous · Answer

He's not that far yet Kira lol

nincompoop · Answer

I wouldn't be hasty using the guy from the Hosptial's Rules... there are other methods you need to learn to get the whole picture of limits. that is the tangent problem

anonymous · Answer

NIce you just got 89 ss nin!

shamil98 · Answer

That's chapter 7 in ma book o.o

nincompoop · Answer

I believe the prior chapters are what we call pre-calculus work out if that were the case

nincompoop · Answer

there are common graphs and equations you need to feel comfortable working with before starting out with calculus.  it is algebra-intensive, so you better be an Algebra Sage or something before you start skipping through concepts

shamil98 · Answer

its paired with integration techniques and improper integrals

nincompoop · Answer

yeah, you're doing the classical method in that case