Why lim x-> 1- of ( x^2 + 2x + 1 ) / (x-1) = - Infinity ?

lim [x→1⁻] (x²+2x+1)/(x−1) = lim [x→1⁻] ((x+1)*(x+1))/(x-1)
When we insert our x(=1⁻) value we get something like this:
((0.999....+1)*(0.999....+1))/(0.999....-1)
Which is actually:
(2*2)/(-0.000....1)
You can see -0.000....1 as 1/(-∞)
So if we make it look nicer we get:
(4/1)/(1/-∞) = (4*(-∞))/1 = -4∞ = -∞
If u didnt understand this step this picture can help you (only the upper part of the photo):

http://tutorial.math.lamar.edu/Classes/Alg/RationalExpressions_files/eq0028P.gif

This isnt really "politically correct" in terms of math but its an easy way to get it
Hope i helped :)

As (x -> 1-) means that X is less than 1 but is increasing toward it.

In those cases, the numerator (x² + 2x + 1) is positive and increasing toward 4, and the denominator is negative and increasing toward 0.

A positive divided by any negative yields a negative result, and as a denominator approaches zero, the quotient tends toward infinity. Hence, the limit here is negative infinity.

Your questions is asking the limit to 1 coming from the negative side of the graph. To prove this, you can plug in numbers slightly less than one, like .999998. This answer is negative and extremely large. So the limit of this function going to one from the negative side is negative infinity.

OMG that is awesome !! I have a test in 40 minutes and those explainations are extremely helpful !! I totally got it.
Thank you guys !!!.

x approaches 1 from the left of 1
lim x-->1- (x^2+2x-1) /(x-1)
The numerator approaches 2 whle the denominator approaches 0 from the left
(denominator is negative while close to 0)
= -∞