Editing Limit of a function (section)

===Limits of compositions of functions===
In general, from knowing that <math>\lim_{y \to b} f(y) = c</math> and <math>\lim_{x \to a} g(x) = b,</math> it does ''not'' follow that <math>\lim_{x \to a} f(g(x)) = c.</math> 
However, this "chain rule" does hold if one of the following ''additional'' conditions holds:
*{{math|1=''f''(''b'') = ''c''}} (that is, {{mvar|f}} is continuous at  {{mvar|b}}), or
*{{mvar|g}} does not take the value  {{mvar|b}} near  {{mvar|a}} (that is, there exists a {{math|''δ'' > 0}} such that if {{math|0 < {{abs|''x'' &minus; ''a''}} < ''δ''}} then {{math|{{abs|''g''(''x'') &minus; ''b''}} > 0}}).

As an example of this phenomenon, consider the following function that violates both additional restrictions:

<math display=block>f(x) = g(x) = \begin{cases}
  0 & \text{if } x\neq 0 \\ 
  1 & \text{if } x=0 
\end{cases}</math>

Since the value at {{math|''f''(0)}} is a [[removable discontinuity]],
<math display=block>\lim_{x \to a} f(x) = 0</math> for all {{mvar|a}}.
Thus, the naïve chain rule would suggest that the limit of {{math|''f''(''f''(''x''))}} is 0.  However, it is the case that
<math display=block>f(f(x))=\begin{cases}
  1 & \text{if } x\neq 0 \\ 
  0 & \text{if } x = 0 
\end{cases}</math>
and so
<math display=block>\lim_{x \to a} f(f(x)) = 1</math> for all {{mvar|a}}.