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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
lim_{x→ a} k = k	lim_{x→ 2} 3 = 3, lim_x→-2 3 = 3
lim_{x→ +∞} k = k	lim_{x→ +∞} 3 = 3, lim_{x→ +∞} 0 = 0
lim_{x→ -∞} k = k	lim_{x→ -∞} 3 = 3, lim_x→-∞ 0 = 0

Now for g(x) = x

Limit

lim_{x→ a} x = a

lim_{x→ +∞} x = +∞

lim_{x→ -∞} x = -∞

lim_{x→ a} k = k

lim_{x→ +∞} k = k lim_{x→ -∞} k = k

(a) y = x

lim_{x→ -∞} x = -∞

(b) y = x

lim_{x→ +∞} x = +∞

(c)

Theorem
Let lim stand for one of the limits
lim_{x→ a}, lim_{x→ a-}, lim_{x→ a+}, lim_{x→ +∞} or lim_{x→ -∞}
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 ⁿ√f(x) = ⁿ√lim f(x) = ⁿ√L₁
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 f_n(x) all exists, then
lim [f₁(x) + f₂ + ... + f_n(x)] = lim f₁(x) + lim f₂(x) + ... + lim f_n(x)
and
lim [f₁(x).f₂(x). ... .f_n(x)] = lim f₁(x).lim f₂(x). ... .lim f_n(x)

If f₁,f₂,………..,f_n 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) = b_nxⁿ + b_{n - 1}x^{n - 1} + ... + b₁x + b₀
Where b_n , b_{n - 1},,…. , b₁ , b₀ are all constants.

Example
lim_{x → 5} (x)² - 4x + 3) = lim_{x → 5} x² - 4lim_{x → 5} x + lim_{x → 5} 3 = (5)² - 4(5) + 3 = 8

Theorem 2.5.2
For any polynomial
p(x) = c₀ + c₁x + ... + c_nxⁿ
and any real number a
lim _{x → a} p(x) = c₀ + c₁a + ... + c_naⁿ = p(a)

Proof
lim_{x → a} p(x) = lim_{x → a} (c₀ + c₁x + ... + c_nxⁿ) = lim_{x → a}c₀ + lim_{x → a}c₁x + ... + lim_{x → a}c_nxⁿ = lim_{x → a}c₀ + c₁lim_{x → a} x + ... + c_nlim_{x → a} xⁿ = c₀ + c₁a + ... + c_naⁿ = p(a)

Limit involving 1/x
The following limits are suggested by the graph of 1/x.

Table of numerical values

	Values	Conslusion
x 1/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.

lim_{x → a-} g(x) = -∞ lim_{x → a+} g(x) = +∞
lim_{x → -∞} g(x) = 0 lim_{x → +∞} g(x) = 0

Limits of polynomiaks as x goes to +∞ and -∞

Lim_{x → +∞} x = +∞ Lim_{x → +∞} x² = +∞

lim_{x → +∞} x³ = +∞

Example
lim_{x → +∞} 2x⁵ = 2lim_{x → +∞} x⁵ = +∞
lim_{x → +∞} -7x⁶ = -7lim_{x → +∞} x⁶ = -∞
For integer value of n
lim_{x → +∞} 1/xⁿ = (lim_{x → +∞} 1/x)ⁿ = 0
lim_{x → -∞} 1/xⁿ = (lim_{x → -∞} 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 c_n = 0 , then
lim_{x → +∞} (c₀ + c₁x + ... + c_nxⁿ) = lim_{x → +∞} c_nxⁿ
lim_{x → -∞} (c₀ + c₁x + ... + c_nxⁿ) = lim_{x → -∞} c_nxⁿ
__________________________________
c₀ + c₁x + ... + c_nxⁿ = xⁿ(c₀/xⁿ + c₁/x^{n - 1} + .. + c_n) = c_nxⁿ

Example
lim_{x → 2} (5x³ + 4)/(x - 3) = [lim_{x → 2} (5x³ + 4)]/[lim_{x → 2} (x - 3)] = [5(2)³ + 4]/[2 - 3] = -44

The graph has not value at x = 2