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I'm studying limits for the first time.

If we graph

$$y = \frac{x^2-1}{x-1},$$

we get a divide by zero 'hole' at $x = 1$ and $y = 2$ in an otherwise straight line. But the above can be factored to

$$y = x + 1$$

which obviates the problem. On the other hand

$$y = \frac{1}{1-x}$$

gives us a hyperbola with the limits at infinity which doesn't seem as stupid as the 'hole' above, and this equation can't be simplified.

My question is: is it always possible to 'repair' my first equation above such that bogus 'limits' can be worked around? I"d like to hope so, since the one 'limit' number of $2$ in the otherwise unbroken line seems very weird. On the other hand, limits at zero and infinity seem quite natural.

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    $\begingroup$ Your first function has what is called a removable discontinuity at $x=1$. And yes, it is always possible to remove removable discontinuities. $\endgroup$ – Zubin Mukerjee Mar 14 '16 at 2:50
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    $\begingroup$ I am hesitant to mention this since it seems you are early in your study of mathematics, but what you are asking is well discussed in the theory of Complex Analysis. The topic of Poles, Zeros, and Removable Singularities should answer this in great detail. $\endgroup$ – JMoravitz Mar 14 '16 at 2:52
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    $\begingroup$ @Rayandrews. Once you get into the concept of a derivative, you will see that this "bogus" limit actually becomes very important, if not a crucial component of where the derivative comes from...Give it some time $\endgroup$ – imranfat Mar 14 '16 at 3:44
  • $\begingroup$ Thanks all. At my level, Zubin's answer is the most 'comfortable' since the 'x = 1, y = 2' situation is indeed what one might well call a discontinuity. If one can always remove these things from otherwise simple line graphs then I'm content. Derivatives ... later. BTW I'm guessing that what you find at a discontinuity is a singularity, no? $\endgroup$ – Ray Andrews Mar 14 '16 at 16:34

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