Question 977471
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The graph is a straight line that has the same solution set as *[tex \Large y\ =\ x\ -\ 1] except for the discontinuity, "hole" if you will, at the point *[tex \Large (-1, -2)] because the denominator of the original function is zero when *[tex \Large x\ =\ -1].


The original function is continuous at *[tex \Large x\ =\ 1] and *[tex \Large \lim_{x\right 1}\,\frac{x^2\ -\ 1}{x\ +\ 1}] 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:


<i>Let *[tex \Large f(x)] be a function defined on an interval that contains *[tex \Large x\ =\ a], <u>except possibly at *[tex \Large \under {x\ =\ a}],</u> then 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right a}\,f(x)\ =\ L] 


if *[tex \Large \forall\ \epsilon\ \in\ \mathbb{R}\ >\ 0\ \exists\ \delta\ \in\ \mathbb{R}\ >\ 0] such that *[tex \Large |f(x)\ -\ L|\ <\ \epsilon] whenever *[tex \Large |x\ -\ a|\ <\ \delta]</i>


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
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \