1. If `lim_(x->c) f(x) g(x) ` exists, then we can have the following cases:


`quadquadquadquad (a)` Both `lim_(x->c)f(x)` and `lim_(x->c) g(x)` exist. Obviously, then `lim_(x->c) f(x) g(x)` exists.


`quadquadquadquad (b)` `lim_(x->c) f(x)` exists and `lim_(x->c)g(x)` does not exist.

Consider `f(x) = x; g(x) = 1/x`, now `lim_(x->0)f(x) *g(x)` exists `=1`. Also `lim_(x->0)f(x) =0` exists but

`lim_(x->0)g(x)` does not exist.


`quadquadquadquad (c)` Both `lim_(x->c)f(x) ` and `lim_(x->c)g(x)` do not exist.

Let `f` be defined as `f(x) = {tt( (1, text(if) x le 0) ,(2, text(if) x >0) )` . Let `g(x) ={tt( (2, text(if) x le 0) ,(1, text(if) x>0) )`

Then `f(x) g(x) = 2`, and so `lim_(x->0) f(x) xx g(x)` exists, while `lim_(x->0)f(x)` and `lim_(x->0)g(x)` do not exist.

2. If `lim_(x->c) [ f(x) +g(x)]` exists then we can have the following cases:

`quadquadquadquad (a)` If `lim_(x->c) f(x)` exists, then `lim_(x->c)g(x)` must exist.

`text( Proof: )` This is true as `g = (f+ g) - f`.

Therefore, by the limit theorem, `lim_(x->c) g(x) =lim_(x->c) (f(x) +g(x)) - lim_(x->0)f(x)` which exists.

`quadquadquadquad (b)` Both `lim_(x->c)f(x) ` and `lim_(x->c)g(x)` do not exist.

Consider `lim_(x->1) [x]` and `lim_(x->1)g[x]`, where `[.]` and `{.}` represent greatest integer and fractional part function, respectively. Here both the limits do not exist but `lim_(x->1)[x]+{x} = lim_(x->1) x=1` exists.


`text(For Example:)`

(a) `f(x) =1/sin x ` and `g(x) =1/tan x` at `x=0` `Lim_(x->0) f(x) = DNE`, `Lim_(x->0)g(x) =DNE`

But `Lim_(x->0 ) (1/sinx -1/tanx) =0 =Lim_(x->0) (1-cos x)/sin x`

`quadquadquadquadquadquadquadquad= Lim_(x->0) (2 sin^2 (x/2))/(2 sin (x/2 )cos (x/2)) `

`quadquadquadquadquadquadquadquad= Lim_(x->0) tan (x/2) =0` (exist).




(b) `f(x) = sgn x` and `g (x) = [x]` then `Lim_(x->0) (sgn x + [x])` does not exist

as `Lim_(x->0^+) sgn x + [x] =1` , `Lim_(x->0)sgn x +[x] =-2`

while `Lim_(x->0) f(x) =DNE`, `Lim_(x->0) g(x) =DNE`


(c) `f(x) = [x] and g(x) = {x}; F(x) = [x] {x}`

`Lim_(x->0) [x] *{x} ` does not exist but `Lim_(x->1)[x]{x}` exist and is equal to zero.

(d) If `f(x) = e^( [ x ])` ; `g(x) = e^({x}) ` then `Lim_(x->0) e^ ([x]) * e^( {x}) = Lim_(x->0) e^x` which exist.

`text(Indeterminate forms : )`

lndeterminant forms are `0/0 , oo/oo, oo -oo, 0 xx oo , 1^ oo ,0^0` and `oo^0`


`(i) text( Direct Substitution Method :)`

Consider the following limits : (i) `lim_(x->a)f(x)` (ii) `lim_(x->a) (phi (x))/(psi (x))`

If `f(a) ` and `(phi (a)) /(psi (a))` exist and are fixed real numbers and ` psi (a) ne 0` then we say that `lim_(x->a) f(x) = f(a)` and `lim_(x-a)(phi (x))/(psi(x)) = (phi (a)) /(psi (a))`

`(ii) text( Fractorization Method :)`

Consider `lim_(x->a) (f(x))/(g(x))`

lf by substituting `x =a`, `f(x) /g(x) ` reduces to the form `0/0`, then `(x - a)` is a factor of both `f(x)` and `g(x)`. So,

we first factorize `f(x)` and `g(x)` and then cancel out the common factor to evaluate the limit.

`(iii) text( Rationalization Method :)`

This is particularly used when either the numerator or the denominator or both involve expression consists

of squares roots and on substituting the value of `x` the rational expression takes the form `0/0, oo/oo`.
Following examples illustrate the procedure.

`(iv) text( Evalutation of Algebraic Limit Using Some Standard Limits :)`

Recall the binomial expansion for any rational power
`(1+x)^n =1 +nx + (n(n-1)) /(2!) x^2 + (n(n-1) (n-2)) /(3!) x^3..........`
where `|x| <1`

When `x` is infinitely small (approaching to zero) such that we can ignore higher powers of x, then we have
`(1 + x)^n = 1 + nx` (approximately).

`(v) text (Evaluation of Algebraic Limits at Infinity )`

We know that `lim_(x-> +oo) 1/x =0` and `lim_(x->+oo) 1/x^2=0`

`:.` `lim_(x->oo)f(x) = lim_(y-> 0)f (1/y)`