Mathematics Revision Notes of Limits for NDA

concept of limits

Suppose f(x) is a real -valued function c is a real number.

The expression `lim_(x->c) f(x) = L` means that f(x) can be as close to L

as desired by making x sufficiently close to c. In such a case, we say that limit off,

as x approaches c, is L.

As x approaches 2, f(x) approaches I and hence we have `lim_(x ->2) f(x) =1`

`f(x) = ( x^2 -4)/(x -2) ` tn this case x approaches 2 the limiting value of f(x) is equal to 4 even if f(x) is not defined at x = 2.



Note that this statement is true even if `f(c) != L`. Indeed, the function f(x) need not even be defined at c.

Two examples help illustrate this.

Consider `f(x) = x - 1` as x approaches 2. In this case, f(x) is defined at 2, and it equals its limiting value 1.

Thus, f(x) can be made arbitrarily close to the limit of 4 just by making x sufficiently close to 2.

Left Hand and Right Hand Limits :

-- Let's Consider the values of the functions at the points which are very near to a on the left of `a`.
- If these values tend to definite unique number as `x` tends to `a` from left, then the unique number, so obtained is called the left hand limit of `f(x)` at `x=a` and symbolically we write it as

`f(a^-) = f(a-0) = lim_(x-> a^(-)) f(x)= lim_(h->0) f(a-h)`

- Similarly, right hand limit can be expressed as

`f(a^+) = f(a+0) = lim_(x-> a^+) f(x) = lim_(h->0) f(a+h)`

Please not that if `oo` or `-oo` are not real

`:.` if `f(a^+) = f(a^-) = oo`

► limit Does Not Exist (DNE) as both LHL & RHL don't exist

Existence of Limit :

`lim_(x-> a) f(x)` exists when

(i) `lim_(x-> a^(-) f(x)` and `lim_(x-> a^+) f(x)` exist i.e. LHL and RHL both exist.

(ii) `lim_(x-> a^(-)) f(x) = lim_(x-> a^+) f(x)` i.e., LHL = RHL

i.e.

`f(a^+) = f(a+0) = lim_(x-> a^+) f(x) = lim_(h->0) f(a+h)`

` = f(a^-) = f(a -0 ) = lim_(x -> a^- ) f(x) = lim_(h ->0) f(a -h)`

Fundamental Theorems on Limits :

If `lim_(x-> a) f(x) = l` and `lim_(x-> a) g(x) = m` (where, `l` and `m` are real numbers), then

(i) `lim_(x-> a) {f(x) +g(x)} =l+m \ \ \ \ " [sum rule]"`

(ii) `lim_(x-> a) {f(x) - g(x) }= l-m \ \ \ \ "[difference rule]"`

(iii) `lim_(x-> a) {f(x) * g(x)} = l* m \ \ \ \ "[product rule]"`

(iv) `lim_(x-> a) k * f(x) = k * l \ \ \ \ "[constant multiple rule]"`

(v) `lim_(x-> a) (f(x))/(g(x))= l/m , m ne 0 \ \ \ \ "[ quotient rule]"`

♦ Imp (vi) If `lim_(x-> a) f(x) = oo` or `-oo`, then `lim_(x-> a) 1/(f(x))=0`

♦ Imp (vii) `lim_(x-> a) log {f(x)}= log {lim_(x-> a) f(x)}`

♦ Imp (viii) If `f(x) le g(x) , AA x` then `lim_(x-> a) f(x) le lim_(x-> a) g(x)`

♦ Imp (ix) `lim_(x-> a) {f(x)}^(g(x)) ={lim_(x-> a) f(x) }^(lim_(x-> a) g(x))`

(x) `lim_(x-> a) f{g(x)} = f{ lim_(x-> a) g(x)}= f(m)` provided `f` is continuous at `lim_(x-> a) g(x) =m`

♦ Imp (xi) `lim_(x -> a) | f(x) | = | lim_(x ->a) f(x) |`

♦ Imp (xii) `lim_(x -> 0 ) (1 + x) ^(1/x) = e`

Some Important Results :

1. Trigonometric Limits :

`(i). lim_(x->0) \ sinx/x = lim_(x->0) \ x/sinx = lim_(x->0) \ sin^(-1)x/x = lim_(x->0) \ x/sin^(-1)x = 1`

`(ii). lim_(x->0) \ tanx/x = lim_(x->0) \ x/tanx = lim_(x->0) \ tan^(-1)x/x = lim_(x->0) \ x/tan^(-1)x = 1`

`(iii). lim_(x->oo) \ sinx/x = lim_(x->oo) \ sin^(-1)x/x = lim_(x->oo) \ tanx/x = lim_(x->oo) \ tan^(-1)x/x = 0`


2. Logarithmic Limits :

`log (1+x)= x-x^2/2 +x^3/3-......` where `-1 < x le 1` and expansion is true only, if base is `e`

3. Exponential Limits :

`e^x =1+x+x^2/(2!) +x^3/(3!)+...............`



Some Usefull Exapansion of function :

(i) `a^x = 1+(x ln a)/(1!) + (x^2 ln^2 a)/(2!) +(x^3 ln^3 a)/(3!) +................ a > 0`

(ii) `e^x = 1+x/(1!) +x^2/(2!) +x^3/(3!) +.................. ∀ x in R`

(iii) `ln (1+x) = x - x^2/2+x^3/3 -x^4/4+ .................... ` for `-1 < x le 1`

(iv) `sinx = x - x^3/(3!) +x^5/(5!) - x^7/(7!) + .................. ( - pi/2 < x < pi/2)`

(v) `cosx = 1 - x^2/(2!) +x^4/(4!) -x^6/(6!) + ........................ x in (-pi/2 , pi/2) `

(vi) `tanx = x+x^3/3+(2x^5)/15+ .......................x in ( -pi/2 , pi/2) `

(vii) `tan^(-1)x = x - x^3/3 +x^5/5-x^7/7 +.....................`

L'Hospital Rule

- It is used to circumvent the common indeterminate forms `0/0` and `∞/∞` when computing limits.

- Suppose that `lim_(x→a) f(x)=0 or oo,` `lim_(x→a)g(x)=0 or oo,` and that functions `f` and `g` are differentiable on an open interval `I` containing `a.`
Assume also that `g′(x)≠0` in `I` if `x≠a` Then

`lim_(x->a) (f(x)) / (g(x)) = lim_(x->a) (f'(x)) / (g'(x)) `

so long as the limit is finite, +∞, or −∞. Similar results hold for `x→∞` and `x→−∞`

 
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