Revision: Calculus >> Limits and Derivatives Mathematics ISC (Commerce) Class 11 CISCE

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. 

Theorem :  Let f and g be two functions such that their derivatives are defined in a common domain. Then

(i) Derivative of sum of two functions is sum of the derivatives of the functions. 
`d/(dx)`[f(x) + g(x)] = `d/(dx)` f(x) + `d/(dx)`

(ii) Derivative of difference of two functions is difference of the derivatives of the functions. 
`d/(dx)` [ f(x) - g(x)] = `d/(dx)` f(x) - `d/(dx)` g(x).

(iii) Derivative of product of two functions is given by the following product rule. 
`d/(dx)`[f(x) . g(x)] = `d/(dx)` f(x) . g(x) + f(x) .`d/(dx)` g(x)

(iv) Derivative of quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero).
`d/(dx)(f(x)/g(x))` =`[d/(dx) f(x) . g(x) - f(x) d/(dx) g(x)]/((g(x))^2)`

Theorem : Let f(x) = `a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0`  be a polynomial function, where `a_is`  are all real numbers and an ≠ 0. Then, the derivative function is given by
`(df(x))/(dx)` = `na_nx^(n-1) a_(n-1)x^(x-2) + ... + 2a_2x + a_1`.

Derivative of trignometric function

1. `d/(dx)` sin x = cos x 

2. `d/(dx)` cos x = - sin x

3.  `d/(dx)` tan x = `sec ^2 x `

4. `d/(dx)` sec x = sec x tan x 

5. `d/(dx)`  cosec x = - cosec x cot x

6. `d/(dx)`  cot x =` - cosec^2 x`

7. `d/(dx)`  log x = `1/ x`

8. `d/(dx)` constant = 0

9. `d/(dx) x^n = n x ^(n - 1)`