Let f ( x ) = p ( x ) / q ( x ) be a rational function. Let q have a root at some x=a. It is clear that if q ( x ) divides p ( x ) then the limit of f as x goes to a exists. But is the converse true as well?

Question

Let   f  (  x  )  =  p  (  x  )      /    q  (  x  ) be a rational function. Let q have a root at some x=a. It is clear that if   q  (  x  ) divides   p  (  x  ) then the limit of   f as   x goes to a exists. But is the converse true as well?

Madisyn Avery · Accepted Answer

The fact that   q  (  x  ) divides   p  (  x  ) cannot be derived if      lim          x      →      a                  p      (      x      )              q      (      x      )      exists, be   a a root of   q  (  x  ) or not.Obviously, if   a is not a root of   q  (  x  ) the limit exists, by continuity.If   a is a root of   q  (  x  ), then you can write   q  (  x  )  =  (  x  −  a      )    m        q    1    (  x  ), where   a is not a root of       q    1    (  x  ): just divide by   x  −  a as many times as you can.In this case   m  &amp;gt;  0 by assumption. You can also write   p  (  x  )  =  (  x  −  a      )    n        p    1    (  x  ), with   a not a root of       p    1    (  x  ); here it is possible that   n  =  0 (when   a is not a root of   p  (  x  ) to begin with. Then we have            p      (      x      )              q      (      x      )        =            (      x      −      a              )        n                    (      x      −      a              )        m                                p        1            (      x      )                      q        1            (      x      )      and, since       q    1    (  a  )  ≠  0, the limit      lim          x      →      a                          p        1            (      x      )                      q        1            (      x      )      exists finite. So we just need to look at                    (*)                              lim                      x            →            a                                                (            x            −            a                          )              n                                            (            x            −            a                          )              m                                          This exists finite if and only if   n  ≥  m, because otherwise it can be rewritten as      lim          x      →      a            1          (      x      −      a              )                  m          −          n                    which does not exist if   m  −  n is odd (it is ∞ from the right and   −  ∞ from the left) and is   ∞ if   m  −  n is even.If   n  =  m, then the limit (  *) is 1; if   n  &amp;gt;  m, then the limit (  *) is 0.Just play with some rational function to understand what happens:      lim          x      →      1                          x        2            +      x      −      2                      x        2            −      1        =  2but obviously   (  x  −  1  )  (  x  +  1  ) does not divide   (  x  −  1  )  (  x  +  2  ).If you instead mean that the limit exists finite for all roots of   q  (  x  ), then the statement is false nonetheless. For instance      lim          x      →      1                          x        2            −      1                      x        3            −              x        2            +      x      −      1      exists finite, and 1 is the only (real) root of       x    3    −      x    2    +  x  −  1

Isaiah Farrell · Accepted Answer

The converse is not necessarily true.For example      lim          x      →      2                          x        2            −      4                      x        2            −      5      x      +      6        =  −  4But   (      x    2    −  5  x  +  6  ) does not divide   (      x    2    −  4  )  .

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