Solve Geometric Limits

geometric limits

In math, the model of a "limit" is used to illustrate the value that a function or order "approaches" as the input or key approaches a number of values. The model of limit allows one to, in a total space; define a new point since a Cauchy sequence of previously clear points. In this article we study about geometric limits and develop the knowledge of the geometry.
Proof for Solve Geometric Limits:

Prove and solve that `lim_(A->0)` ` (sin A)/ A` = 1




geometric limit

We take y =` (sin A)/A`  . This function is defined for all A, other than A = 0, for which both numerator and denominator become zero. When A is replaced by − A, the magnitude of the fraction`(sin A)/A` does not change since `(sin (-A))/(- A)` =`(sin A)/A .`

Therefore it is enough to find the limit of the fraction as θ tends to 0 through positive values. i.e. in the first quadrant. We consider a circle with centre at O radius unity. P, Q are two points on this circle so

OP = OQ = 1. Let A be the angle subtended at the centre by the arc PT. Measuring angle in radians, we have sin A = PR, R being a point on PR such that OS passes through R.  cosA = OR,P  =`1/2` arc PQ, OPS = 90°

In triangle OPS, PS = tanA. Now length of arc AB = 2A and length of the chord AB = 2 sinA sum of the tangents = PS + QS = 2 tan A

Since the length of the arc is intermediate between the length of chord and the sum of the tangents we can write 2 sin A < 2A < 2 tan A.

Dividing by 2 sin A , we have 1 < A sinA <`1/(cos A)` or 1 >`(sin A)/A ` > cos A . But as A → 0, cos A, given by the distance OR, tends to 1

That is, `lim_(A->0)` cos A = 1 . Therefore 1 >`lim_(A->0)``( sin A)/A` > 1, by 3 .

That is, the variable y =`(sin A)/A` always lies between unity and a magnitude tending to unity, and hence `lim_(A->0)` `(sin A)/A` = 1.


Examples for Solve Geometric Limits:

Solve that using geometric limits `lim_(A->0)` `(1 - cos A)/A^2` .


`(1 - cosA)/A^2`    = `{2(sin^2 A/2)}/(A^2)`

`(1 - cosA)/A^2`     =  `{(1/2){ sin^2(A/2)}/(A/2)^2}`

=`1/2`    `{ sin(A/2)/(A/2)}^2`

If A→ 0, α =`A/2` also tends to 0 and `lim_(A->0)`  `sin(A/2)/(A/2)`  =limα → 0 ( sin α)/α = 1 and

hence`lim_(A->0)` ` (1- cosA)/A^2` =`lim_(A->0)` ` ((1/2)(sin(A/2))/(A/2))^2`

=`1/2` `lim_(A->0)`  `{ sin(A/2)/(A/2)}^2`

=`1/2` × `1^2 ` =`1/2`