Revision: 11th Std >> Limits MAH-MHT CET (PCM/PCB) Limits

Theorem: 
 Let f  and g be two functions such that both `lim_(x -> a)` f(x) and `lim_(x -> a)` g(x) exist.
Then

(i) Limit of sum of two functions is sum of the limits of the functions, i.e.,
`lim_(x -> a) [f(x) + g(x)]` = `lim_(x -> a) f(x) + lim _(x -> a) g(x)`.

(ii) Limit of difference of two functions is difference of the limits of the functions, i.e.,
`lim_(x -> a) [f(x) -g(x)] = lim_(x -> a) f(x) -lim _(x -> a) g(x)`.

(iii) Limit of product of two functions is product of the limits of the functions, i.e.,
`lim_(x -> a) [f(x)  . g(x)] = lim_(x -> a) f(x)  .  lim _(x -> a) g(x)`.

(iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e., 
`lim_(x -> a) f(x)/g(x)  =(lim_(x->a) f(x))/(lim_(x-> a )  g(x))`

Theorem : (Sandwich Theorem)

Let f, g and h be real functions such that f (x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a , if
`lim_(x - > a)`f(x) = l = `lim_(x ->a)` h(x) , then `lim_(x ->a)` g(x) = l .fig.

Given below is a beautiful geometric proof of the following important inequality relating trigonometric functions.
`cos x < sin x/x < 1  for 0 <|x| < pi / 2`
Proof :   We know that sin (– x) =  – sin x and cos( – x) = cos x. Hence, it is sufficient  to prove the inequality for 0 < x < `pi /2`. In the following fig.


O is the centre of the unit circle such that  the angle AOC is x radians and 0 < x <`pi /2` .Line segments B A and  CD are perpendiculars to OA. Further, join AC. Then
Area of OAC ∆ < Area of sector OAC < Area of ∆ OAB .

i.e.,`1/2`OA.CD <`x / 2pi` .`( OA) ^2` < `1/2` OA .AB.
i.e., CD < x . OA < AB.
From ∆ OCD,
sin x = `(CD)/(OA)` (since OC = OA)   and hence CD = OA sin x. Also  tan x = `(AB)/(OA)` and hence  AB = OA. tan x. 
Thus OA sin x < OA. x < OA. tan x. 
Since length OA is positive, 
we have sin x < x < tan x.
Since 0 < x < `pi/2` , sinx is positive and thus by dividing throughout by sin x, we have `1 < x/(sin x) < 1/(cos x)` .  Taking reciprocals throughout , we have
`cos x <  (sin x)/x < 1`
which complete the proof.

Theorem - The following are two important limits. 
i) `lim_(x -> 0) sin x / x = 1` 

ii) `lim_(x->0) (1 - cos x ) / x = 0`

Proof : 
i) The  inequality in (*) says that the function `sin x / x ` is sandwiched between the functions cos x and the constant function which takes value 1 . 
Further, since `lim _(x→0)` cos x = 1, we see that the proof of (i) of the theorem is complete by sandwich theorem.

ii)  we recall the trigonometric identity
1 – cos x = 2 `sin^2 (x / 2)`
Then 

`lim_(x -> 0) (1 - cos x)/x = lim_(x -> 0) 2 sin^2 (x/2) / x = lim_(x-> 0) sin (x / 2) / (x / 2) . sin (x / 2)`

`lim_(x -> 0) sin(x / 2) / (x / 2) . lim_(x->0) sin (x/2) = 1.0 = 0`

Observe that we have implicitly used the fact that 0 x → is equivalent to `x/2 -> 0` . This may be justified by putting  `  y = x / 2`    

Theorem :
 Let f and g be two real valued functions with the same domain such that f(x) ≤ g(x)  for all x in the domain of definition, For some a, if both
`lim_( x -> a)` f(x) and `lim _(x - >a)` g(x) exist ,
then
`lim_(x ->a)` f(x) ≤ `lim_(x -> a)` g(x).
The explain in following fig.