Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updates: 27 May 2012


l= limxa f(x)    means:    If ε >0 then there exists δ>0 such that if d(x,a) <0 then d(f(x),l) <ε.

l= limn an    means:    If ε >0 then there exists N >0 such that if n >0 and n>N then d(an,l)<ε.

Let X be a metric space. Let f:X and g:X be functions and let aX. Assume that limxa f(x) and limxa g(x) exist. Then

  1. (a) limxa (f(x)+g(x) )= limxa f(x)+ limxa g(x),
  2. (b) If c then limxa cf(x) =climxa f(x),   and
  3. (c) limxa f(x)g(x) =(limxa f(x)) (limxa g(x)) .

Proof of part (a).

Proof of part (b):

Proof of part (c):

Assume that limxa g(x)=l and limyl f(y) exists. Then limyl f(y)= limxa f(g(x)).


Let an and bn be sequences in . Assume that limn an exists and limn bn exists. If anbn then limn an limn bn.


The functions f(x)=xn and f(x)=ex

  1. (a) Let n >0 and a. Then xn is continuous at x=a (i.e. limxa xn=an).
  2. (b) ex is continuous at x=a.
  3. (c) Prove that limx0 ex=e0.
  4. (d) Prove that limxa ex=ea.

Proof of part (a).

First part of proof of part (b).

Proof of part (d).

  1. (a) Let x. limn xn ={ 0, if |x|<1, diverges, if |x|>1, 1, ifx=1, diverges, if |x|=1 and x1,
  2. (b) Let x. limn 1+x+x2+ +xn = limn 1-xn+ 11-x ={ 11-x, if |x|<1, diverges, if|x| 1.

Proof of part (a) Case |x|<1.

Proof of part (a) Case |x|>1.

Proof of part (b).

Useful limits

  1. (a) If n >0 then limx xne-x =0.
  2. (b) If α >0 then limx x-α logx=0.
  3. (c) Let p >0. Then limn 1np =0 .
  4. (d) Let p >0. Then limn p1/n =1 .
  5. (e) limn n1/n =1.
  6. (f) Let α and p >0. Then limn nα (1+p)n =0 .
  7. (g) If |x|<1 then limn xn=0.
  8. (h) If 0<α<1 then limx0 xα=0.
  9. (i) limx0 ex-1x =1.
  10. (j) limx0 sinxx=1 .
  11. (k) limx0 cosx-1 x2 =-12 .
  12. (l) limx0 log(1+x) x=1 .

Proof of part (l):
limx0 log(1+x) x = limy0 log(1+ ey-1) ey-1 = limy0 yey-1 = limy0 1 ( ey-1 y ) = 11 = 1.

Proof of part (j):
limx0 sinxx = limx0 eix -e-ix 2ix = limx0 eix-1 2ix - e-ix-1 2ix = limx0 12 ( eix-1 ix ) + 12 ( e-ix-1 -ix ) = 12·1 +12·1 = 1.

Proof of part (a):
Assume n >0. 0 limx xne-x = limx xnex limx xn 1 (n+1)! xn+1 = limx (n+1)!x = (n+1)! limx 1x = 0.

Proof of part (b):
Let 0<ε<α. 0 limx x-α logx = limx (x-α 1x 1tdt) limx (x-α 1xt ε-1dt) = limx x-α ( xε-1ε ε ) = limx ( xε-α -x-α ε) limx xε-α ε = 0.

Proof of part (c).
To show: If ε >0 then there exists N >0 such that if n >0 and n>N then | 1np | <ε .
Assume ε >0.
Let N= (1ε) 1p .
To show: If n >0 and n>N then |1np| <ε .
Assume n >0 and n>N.
To show: |1np| <ε .
|1np| < |1Np| = 1( (1ε) 1p) p = 1(1ε) = ε.

Proof of part (d):

Proof of part (e):

Notes and References

This section proves the fundamental limit theorems used most often for computations. THe presentation is focused on Theorem 1.1, establishing, without much effort, the primary tools that are used for evaluation of limits in examples.

It feels a bit funny not to include the f(x)/g(x) case in Theorem 1.1. SHOULD IT BE PUT IN? WHY? WHY NOT?

It is interesting to note that in practice, once Theorems 1.1 and 1.2 are proved, it is hardly ever necessary to use an ε-δ proof to evaluate a limit.

A traditional reference for this material is [Ru1, Ch. 3 and 4] with Theorem 1.1 being [Ru1, Theorem 4.4] and much of Theorem ??? coinciding with [Ru1, Theorem 3.20].


[Bou] N. Bourbaki, Algèbre, Chapitre ?: ??????????? MR?????.

[Ru1] W. Rudin, Principles of mathematical analysis, McGraw-Hill, 1976. MR0385023

[Ru2] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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