Monday, August 22, 2016

Unizor - Derivatives - Indeterminate Forms

Notes to a video lecture on

Sequence Limit -
Indeterminate Forms

When a sequence is represented by a short simple formula of the order number n, like {1/n²}, it's easy to find its limit.

When a sequence is a simple operation (sum or product) on other sequences with known or easily obtainable limits, like
it's easy too - just perform the operation on corresponding limits.

The problem arises when we cannot break a sequence into individual components, determine a limit for each component and do the required operations on the limits. Here are a few simple examples:
(a) Xn = (1/n)·(2n+3)
here 1/n is infinitesimal and 2n+3 is infinitely growing, their product is undefined.
(b) Xn = (2n+3)/n
here both numerator and denominator are infinitely growing with undefined limit (you may say that limit is infinity, but it's not a number, so you cannot perform an operation of division anyway)
(c) Xn = sin(n)/n²
here numerator is not even a convergent sequence, while denominator is infinitely growing.

In all cases where we cannot simply determine the limit of a complex sequence as a result of a few operations on the limits of the components, we deal with indeterminate forms.
Each such case should be dealt in some way, very specific for each given sequence, to transform the sequence into equivalent, but easier to deal with form.

Let's consider different cases we might have using concrete example of sequences.

1. Ratio of two infinitesimals
(indeterminate of type 0/0)

(a) (2−(n+1)+3−n)/2−n =
2−1+(3/2)−n → 1/2

(b) sin²(1/n) / [1−cos(1/n)] =
= sin²(1/n)·[1+cos(1/n)] /
/ [1−cos²(1/n)] =
= 1+cos(1/n) → 2

2. Product of infinitesimal and infinitely growing sequence
(indeterminate of type 0·∞)

(a) [1/(n+1)]·n² =
(n−1)(n+1)/(n+1)+1/(n+1) =
which is an infinitely growing sequence

(b) n·sin(1/n) =
sin(1/n) / (1/n) → 1
see lecture Trigonometry - Trigonometric Identities and Equations - Geometry with Trigonometry - Lim sin(x)/x, where it was proven that the limit of sin(x)/x as x→0 equals to 1.

3. Ratio of two infinitely growing sequences
(indeterminate of type ∞/∞)

(a) (n²+n−2)/(2n²+3n−5) =
[(n+2)(n−)]/[(2n+5)(n−1)] =
(n+2)/(2n+5) =
[(2n+5)−1]/[2(2n+5)] =
1/2 − 1/[2(2n+5)] → 1/2

(b) [n²·sin(1/n)+1]/n =
n·sin(1/n)+1/n =
sin(1/n)/(1/n)+1/n → 1

4. Difference of two infinitely growing sequences
(indeterminate of type ∞−∞)

(a) (n²+n)−n =
/ [√(n²+n)+n] =
/ [n√(1+1/n)+n] =
/ [√(1+1/n)+1] → 1/2

(b) n+1−log2(2n−1) =
1+log2[2n/(2n−1)] =
1+log2[1+1/(2n−1)] → 1

Many indeterminate forms do have a limit, but, to find it, it's necessary to transform the original sequence into equivalent without indeterminate components.

No comments: