SOLUTION: 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 questio
Algebra ->
Rational-functions
-> SOLUTION: 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 questio
Log On
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? Doesn't the limit have to be approaching an aymptote? Since it is approaching 1, I think y=0. I don't understand how that is possible if x=1 is not a discontinuity. My book says that there is a hole at x=1, but I have tried graphing it on a calculator and cannot seem to find it.
Thank you very much!
You can put this solution on YOUR website! 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?
Look at these two graphs:
The graph on the left is of , the graph on the right is of .
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.
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".
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
The graph is a straight line that has the same solution set as except for the discontinuity, "hole" if you will, at the point because the denominator of the original function is zero when .
The original function is continuous at and is simply the value of the function at 1.
A limit can be taken at any value and that value does not necessarily have to be in the domain of the function. Review the definition of a limit:
Let be a function defined on an interval that contains , except possibly at , then
if such that whenever
I'm quite surprised that the question asked you for the limit at 1, a very trivial and uninteresting question indeed. I wouldn't be surprised to find out that when you went back and checked your assignment that the limit of interest is actually the limit as x goes to -1.
John
My calculator said it, I believe it, that settles it
