Question 977471
Hello! 
How do I graph f(x)=(x^2-1)/(x+1)?
What does lim(x->1) (x^2-1)/(x+1) equal?

I graphed the function with a discontinuity at x=-1, but I'm not sure how my second question relates?
<pre>
Look at these two graphs:

{{{drawing(300,300,-5,5,-5,5,
graph(300,300,-5,5,-5,5), line(-12,-13,12,11) )}}}{{{drawing(300,300,-5,5,-5,5,
graph(300,300,-5,5,-5,5), line(-12,-13,-1.1,-2.1),line(-.9,-1.9,12,11),
green(circle(-1,-2,.15))

 )}}}

The graph on the left is of {{{y = x-1}}}, the graph on the right is of {{{y = f(x) = (x^2-1)/(x+1)}}}.

They are exactly alike except for one point (-1,-2).

Both graphs approach that point from both the left and the right.  The
difference is that the one on the left actually 'gets there' whereas the
one on the right skips over it.  There's a hole in it there.  Another name
for the "hole" is "removable discontinuity".  How do you remove it?
By factoring the numerator and cancelling the denominator. 

{{{(x^2-1)/(x+1)=((x-1)(x+1))/(x+1)=((x-1)(cross(x+1)))/(cross(x+1))=x-1}}}  
 
But in spite of what you were told in algebra courses, that cancellation 
is not permitted except when you are sure that x is NOT -1.  f(x) is not 
defined when x=-1

However the limit does not ask "What is f(-1)?" It can't, for there is no such! 
It asks instead "What number does y or f(x) get extremely close to when x gets
extremely close to -1 either from the right or from the left?".  That answer is
"It gets extremely close to -2". 

{{{matrix(2,2,

lim,(x^2-1)/(x+1), 
"x->1","")}}}{{{""=""}}}{{{matrix(2,2,

lim,((x-1)(x+1))/(x+1), 
"x->1","")}}}{{{""=""}}}{{{matrix(2,2,

lim,((x-1)(cross(x+1)))/(cross(x+1)), 
"x->1","")}}}{{{""=""}}}{{{matrix(2,2,

lim,x+1, 
"x->1","")}}}{{{""=""}}}{{{-2}}}

It's OK to cancel when getting the limit, because both graphs above
approach the same point, one 'gets there' and one doesn't.  But they
have that same limit of -2.  

You have to understand this concept of asking "What is y approaching
when x is approaching a given number?" for it's important in understanding
calculus.  

Edwin</pre>