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The concept of the derivative of a function is what most people think of as the beginning of calculus.However, before we can even define the derivative we must introduce a kind of generalization of the notion of continuity. That is, we must beginwith the definition of the limit of a function.

The concept of the derivative of a function is what most people think of as the beginning of calculus.However, before we can even define the derivative we must introduce a kind of generalization of the notion of continuity. That is, we must beginwith the definition of the limit of a function.

Let f : S C be a function, where S C , and let c be a limit point of S that is not necessarily an element of S . We say that f has a limit L as z approaches c, and we write

lim z c f ( z ) = L ,

if for every ϵ > 0 there exists a δ > 0 such that if z S and 0 < | z - c | < δ , then | f ( z ) - L | < ϵ .

If the domain S is unbounded, we say that f has a limit L as z approaches , and we write

L = lim z f ( z ) ,

if for every ϵ > 0 there exists a positive number B such that if z S and | z | B , then | f ( z ) - L | < ϵ .

Analogously, if S R , we say lim x f ( x ) = L if for every ϵ > 0 there exists a real number B such that if x S and x B , then | f ( x ) - L | < ϵ . And we say that lim x - f ( x ) = L if for every ϵ > 0 there exists a real number B such that if x S and x B , then | f ( x ) - L | < ϵ .

Finally, for f : ( a , b ) C a function of a real variable, and for c [ a , b ] , we define the one-sided (left and right) limits of f at c . We say that f has a left hand limit of L at c , and we write L = lim x c - 0 f ( x ) , if for every ϵ > 0 there exists a δ > 0 such that if x ( a , b ) and 0 < c - x < δ then | f ( x ) - L | < ϵ . We say that f has a right hand limit of L at c , and write L = lim x c + 0 f ( x ) , if for every ϵ > 0 there exists a δ > 0 such that if x S and 0 < x - c < δ then | f ( x ) - L | < ϵ .

The first few results about limits of functions are not surprising. The analogy between functions having limits and functions being continuous is very close,so that for every elementary result about continuous functions there will be a companion result about limits of functions.

Let c be a complex number. Let f : S C and g : S C be functions. Assume that both f and g have limits as x approaches c . Then:

  1. There exists a δ > 0 and a positive number M such that if z S and 0 < | z - c | < δ then | f ( z ) | < M . That is, if f has a limit as z approaches c , then f is bounded near c .
  2. lim z c ( f ( z ) + g ( z ) ) = lim z c f ( z ) + lim z c g ( z ) .
  3. lim z c ( f ( z ) g ( z ) ) = lim z c f ( z ) lim z c g ( z ) .
  4. If lim z c g ( z ) 0 , then
    lim z c f ( z ) g ( z ) = lim z c f ( z ) lim z c g ( z ) ,
  5. If u and v are the real and imaginary parts of a complex-valued function f , then u and v have limits as z approaches c if and only if f has a limit as z approaches c . And,
    lim z c f ( z ) = lim z c u ( z ) + i lim z c v ( z ) .
  1. Prove [link] . HINT: Compare with [link] .
  2. Prove that lim x c f ( x ) = L if and only if, for every sequence { x n } of elements of S that converges to c , we have lim f ( x n ) = L . HINT: Compare with [link] .
  3. Prove the analog of [link] replacing the limit as z approaches c by the limit as z approaches .
  1. Prove that a function f : S C is continuous at a point c of S if and only if lim x c f ( x ) = f ( c ) . HINT: Carefully write down both definitions, and observe that they are verbetim the same.
  2. Let f be a function with domain S , and let c be a limit point of S that is not in S . Suppose g is a function with domain S { c } , that f ( x ) = g ( x ) for all x S , and that g is continuous at c . Prove that lim x c f ( x ) = g ( c ) .

Prove that the following functions f have the specified limits L at the given points c .

  1. f ( x ) = ( x 3 - 8 ) / ( x 2 - 4 ) , c = 2 , and L = 3 .
  2. f ( x ) = ( x 2 + 1 ) / ( x 3 + 1 ) , c = 1 , and L = 1 .
  3. f ( x ) = ( x 8 - 1 ) / ( x 6 + 1 ) , c = i , and L = - 4 / 3 .
  4. f ( x ) = ( sin ( x ) + cos ( x ) - exp ( x ) ) / ( x 2 ) , c = 0 , and L = - 1 .

Define f on the set S of all nonzero real numbers by f ( x ) = c if x < 0 and f ( x ) = d if x > 0 . Show that lim x 0 f ( x ) exists if and only if c = d .

(b) Let f : ( a , b ) C be a complex-valued function on the open interval ( a , b ) . Suppose c is a point of ( a , b ) . Prove that lim x c f ( x ) exists if and only if the two one-sided limits lim x c - 0 f ( x ) and lim x c + 0 f ( x ) exist and are equal.

Suppose f : S C is a function, and that lim x c f ( x ) = L . Define a function g by g ( y ) = f ( y + c ) .

  1. What is the domain of g ?
  2. Show that 0 is a limit point of the domain of g and that lim y 0 g ( y ) = lim x c f ( x ) .
  3. Suppose T C , that h : T S , and that lim y d h ( y ) = c . Prove that
    lim y d f ( h ( y ) ) = lim x c f ( x ) = L .

REMARK When we use the word “ interior” in connection with a set S , it is obviously important to understand the context; i.e., is S being thought of as a set of real numbers or as a set of complex numbers.A point c is in the interior of a set S of complex numbers if the entire disk B ϵ ( c ) of radius ϵ around c is contained in S . While, a point c belongs to the interior of a set S of real numbers if the entire interval ( c - ϵ , c + ϵ ) is contained in S . Hence, in the following definition, we will be careful to distinguish between the cases that f is a function of a real variable or is a function of a complex variable.

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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