Limits and Computational Approach

Limits and Computational Techniques
• Algebraic techniques for finding limits
• How to use limits of basic function to compute limits of complicated function.

Some basic limit for
f(x) = k       The constant function
g(x) = x       Straight line through origin
f(x) = k

 Limit Example limx→ a k = k limx→ 2 3 = 3, limx→-2 3 = 3 limx→ +∞ k = k limx→ +∞ 3 = 3, limx→ +∞ 0 = 0 limx→ -∞ k = k limx→ -∞ 3 = 3, limx→-∞ 0 = 0

Now for g(x) = x

 Limit limx→ a x = a limx→ +∞ x = +∞ limx→ -∞ x = -∞ limx→ a k = k limx→ + k = k     limx→ - k = k
(a)    y = x limx→ - x = -∞
(b)    y = x limx→ + x = +∞
(c) Theorem
Let lim stand for one of the limits
limx→ a, limx→ a-, limx→ a+, limx→ + or limx→ -
If L1 = lim f(x) and L2 = lim g(x) both exist, then
a) lim [f(x) + g(x)] = lim f(x) + lim g(x) = L1 + L2
b) lim [f(x) - g(x)] = lim f(x) - lim g(x) = L1 - L2
c) lim [f(x).g(x)] = lim f(x).lim g(x) = L1.L2
d) lim [f(x)/g(x)] = [lim f(x)/lim g(x)] = L1/L2
e) lim   nf(x) =   nlim f(x) =   nL1
L1 ≥ 0 if n is even
Remark
Although the results (a) and (c) are stated for two functions f and g, these results hold as well for and finite number of functions; that is, if the limits lim f1(x),   Lim f2(x),……….lim fn(x) all exists, then
lim [f1(x) + f2 + ... + fn(x)] = lim f1(x) + lim f2(x) + ... + lim fn(x)
and
lim [f1(x).f2(x). ... .fn(x)] = lim f1(x).lim f2(x). ... .lim fn(x)
If f1,f2,………..,fn are same functions
lim [f(x)]n = [lim f(x)]n
Thus we can write
A) lim x → a xn = [lim x → a x]n = an
Another useful result
B) lim [k.g(x)] = lim (k).lim g(x) = k.lim g(x)
Where k is constant
Polynomial
A polynomial is an expression of the form
f(x) = bnxn + bn - 1xn - 1 + ... + b1x + b0
Where bn , bn - 1,,…. , b1 , b0 are all constants.
Example
limx → 5 (x)2 - 4x + 3) = limx → 5 x2 - 4limx → 5 x + limx → 5 3 = (5)2 - 4(5) + 3 = 8
Theorem 2.5.2
For any polynomial
p(x) = c0 + c1x + ... + cnxn
and any real number a
lim x → a p(x) = c0 + c1a + ... + cnan = p(a)
Proof
limx → a p(x) = limx → a (c0 + c1x + ... + cnxn) = limx → ac0 + limx → ac1x + ... + limx → acnxn = limx → ac0 + c1limx → a x + ... + cnlimx → a xn = c0 + c1a + ... + cnan = p(a)
Limit involving 1/x
The following limits are suggested by the graph of 1/x. Table of numerical values
 Values Conslusion x1/x 1 .. 0.1 .. 0.001 ..1 .. 100 .. 1000 .. x → 0+1/x → +∞ x 1/x -1 .. -0.01 .. -0.001-1 .. -100 .. -1000 .. x → 0-1/x → -∞ x 1/x 1 .. 100 .. 1000 .. 1 .. 0.01 .. 0.001 .. x → +∞1/x decreases towards 0 x 1/x -1 .. -100 .. -1000 ..-1 .. -0.01 .. -0.001 .. x → -∞1/x increases towards 0

For every real number a the graph of the function g(x) = 1/(x - a) is a translation of f(x) = 1/x by a units to the right. limx → a- g(x) = -∞     limx → a+ g(x) = +∞
limx → - g(x) = 0     limx → + g(x) = 0
Limits of polynomiaks as x goes to +∞ and -∞  Limx → + x = +∞     Limx → + x2 = +∞ limx → + x3 = +∞
Example
limx → + 2x5 = 2limx → + x5 = +∞
limx → + -7x6 = -7limx → + x6 = -∞
For integer value of n
limx → + 1/xn = (limx → + 1/x)n = 0
limx → - 1/xn = (limx → - 1/x)n = 0 y = f(x) = 1/xn (n is a positive odd integer) A polynomial behaves like its term of highest degree as x→ +∞ or x→ -∞ more precisely, if cn = 0 , then
limx → + (c0 + c1x + ... + cnxn) = limx → + cnxn
limx → - (c0 + c1x + ... + cnxn) = limx → - cnxn
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c0 + c1x + ... + cnxn = xn(c0/xn + c1/xn - 1 + .. + cn) = cnxn
Example
limx → 2 (5x3 + 4)/(x - 3) = [limx → 2 (5x3 + 4)]/[limx → 2 (x - 3)] = [5(2)3 + 4]/[2 - 3] = -44
The graph has not value at x = 2 Example
limx → 4+ (2 - x)/(x - 4)(x + 2) limx → 4+ (2 - x)/(x - 4)(x + 2) = -∞
Quick method for finding limit of rational functions Contact email: 