Revision: Limit, Continuity, and Differentiability JEE Main Limit, Continuity, and Differentiability

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 :
 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 : (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`    

Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.

f(x) = |x − 1|, x ∈ R

f(x) = (x − 1), if x − 1 > 0

= −(x − 1), if x − 1 < 0

At x = 1

f(1) = 1 − 1 = 0

left-side limit:

`lim_(h -> 0^-) (f(1 - h) - f(1))/ -h`

= `lim_(h -> 0^-) (1 - (1 - h) - 0)/ (- h)`

= `lim_(h -> 0^-) (+ h)/(- h)`

= −1

Right-side limit:

= `lim_(h -> 0^+) (f(1 + h) - f(1))/h`

= `lim_(h -> 0^+) ((1 + h) - 1 - 0)/ h`

= `lim_(h -> 0^+) h/h`

= 1

Left-side limit and the right-side limit are not equal.

Hence, f(x) is not differentiable at x = 1.

Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.

f(x) = [x], 0 < x < 3

(i) At x = 1

Left-side limit:

`lim_(h -> 0) ([1 - h] - [1])/-h`

= `lim_(h -> 0) (0 - 1)/-h`

= `lim_(h -> 0) 1/h`

= Infinite (∞)

Right-hand limit:

`lim_(h -> 0) ([1 + h] - [1])/h`

= `lim_(h -> 0) (1 - 1)/h`

= 0

Left-side limit and right-side limit are not equal.

Hence, f(x) is not differentiable at x = 1.

(ii) At x = 2

Left-side limit:

`lim_(h -> 0) (f(2 + h) - f(2))/h`

= `lim_(h -> 0) ([2 + h]-2)/h`

= `lim_(h -> 0) (2 -2)/h`

= 0

Right-hand limit:

`lim_(h -> 0) (f(2 - h) - f (2))/h`

= `lim_(h -> 0) ([2 - h] - [2])/-h`

= `lim_(h -> 0) (1 - 2)/-h`

= Infinite (∞)

Left-side limit and right-side limit are not equal.

Hence, f(x) is not differentiable at x = 2.

y = `|(f(x), g(x), h(x)),(l, m, n),(a, b, c)|`

`dy/dx= |(d/dx (f(x)), d/dx (g(x)), d/dx (h(x))), (l, m, n), (a, b, c)| + |(f(x), g(x), h(x)),(0, 0, 0),(a, b, c)| + |(f(x), g(x), h(x)),(l, m, n),(0, 0, 0)|`

`= |(f'(x), g'(x), h'(x)),(l, m, n),(a, b, c)|`

Given, (x – a)² + (y – b)² = c²  ...(1)

On differentiating with respect to x,

`=> 2 (x - a) + 2(y - b) dy/dx = 0`

`=> (x - a) + (y - b) dy/dx = 0`  ...(2)

Differentiating again with respect to x,

`1 + dy/dx * dy/dx + (y - b) (d^2 y)/dx^2` = 0

`1 + (dy/dx)^2 + (y - b) (d^2y)/dx^2` = 0

`=> (y - b) = - {(1 + (dy/dx)^2)/((d^2y)/dx^2)}`  ...(3)

Putting the value of (y – b) in (2),

`(x - a) = {(1 + (dy/dx)^2)/((d^2y)/dx^2)}(dy/dx)`   ...(4)

Putting the values ​​of (x − a) and (y − b) from (3) and (4) in (1),

`{1 + (dy/dx)^2}^2/((d^2y)/dx^2)^2 * (dy/dx)^2 + {(1 + (dy/dx)^2)/((d^2y)/dx^2)} = c^2`

On multiplying by `((d^2y)/dx^2)^2`,

`[1 + (dy/dx)^2]^2 (dy/dx)^2 + [1 + (dy/dx)^2]^2 = c^2 ((d^2y)/dx)^2`

`=> [1 + (dy/dx)^2]^2 [(dy/dx)^2 + 1] = c^2 ((d^2y)/dx^2)^2`

`=> {1 + (dy/dx)^2}^3 = c^2 ((d^2y)/dx^2)^2`

On taking the square root,

`therefore {1 + (dy/dx)^2}^(3//2)/((d^2y)/dx^2)` = c  ...(a constant independent of a and b.)

`x sqrt(1 + y) + y sqrt(1 + x) = 0`

∴ `xsqrt(1 + y) = - y sqrt(1 + x) = 0`

On squaring both sides,

x² (1 + y) = y² (1 + x)

⇒ x² + x²y = y² + y²x

⇒ x² – y² – y²x + x²y = 0

⇒ (x – y)(x + y) + xy(x – y) = 0

⇒ (x – y)[x + y + xy] = 0

x – y = 0

As x ≠ y

x + y (1 + x) = 0

y (1 + x) = –x

∴ y = `-x/(1 - x)`

Differentiating w.r.t., x

∴ `dy/dx = ((1 + x)(1) - x (1))/(1 + x)^2`

= `-(1 + x - x)/(1 + x)^2`

= `-1/(1 + x)^2`

`x sqrt(1 + y) + y sqrt(1 + x) = 0`

∴ `xsqrt(1 + y) = - y sqrt(1 + x) = 0`

On squaring both sides,

x² (1 + y) = y² (1 + x)

⇒ x² + x²y = y² + y²x

⇒ x² – y² – y²x + x²y = 0

⇒ (x – y)(x + y) + xy(x – y) = 0

⇒ (x – y)[x + y + xy] = 0

x – y = 0

As x ≠ y

x + y (1 + x) = 0

y (1 + x) = –x

∴ y = `-x/(1 - x)`

Differentiating w.r.t., x

∴ `dy/dx = ((1 + x)(1) - x (1))/(1 + x)^2`

= `-(1 + x - x)/(1 + x)^2`

= `-1/(1 + x)^2`

Given, y = 5 cos x – 3 sin x

Differentiating both sides with respect to x,

`dy/dx = 5 d/dx cos x - 3 d/dx sin x`

= 5 (−sin x) − 3 cos x

= −5 sin x − 3 cos x

Differentiating both sides again with respect to x,

`(d^2 y)/dx = - 5 d/dx sin x - 3 d/dx cos x`

= −5 cos x − 3 (−sin x)

= 3 sin x − 5 cos x

Hence, `(d^2 y)/dx^2 + y` = 0

(3 sin x − 5 cos x) + (5 cos x − 3 sin x) = 0 ...(On substituting the value of y)

Statement:

If a function f(x):

1. Is continuous on the closed interval [a,b]

2. Is differentiable on the open interval (a,b)

3. Satisfies f(a) = f(b)

Then there exists at least one $c\in (a,b)$ such that:

\[f^{\prime}(c)=0\]

Statement: 

If a function f(x):

1. Is continuous on the closed interval [a,b]

2. Is differentiable on the open interval (a,b)

Then there exists at least one number c ∈ (a,b) such that:

\[f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\]