Theorem: Let f(x)→L, g(x)→L' then x→a and cεR  then we have
(a) f(x)+g(x)→L+L'
(b) cf(x)→cL
(c) f(x)g(x)→LL' as x→a
(d) if further L'≠0 then g(x)≠0 for all x→a and distinct from a,  f(x)/g(x)=L/L'.

Proof : 

(a)        we are going to proof our argument by using the concept of convergent functions.
Now if ε>0 then there exists δ>0 such that |f(x)-g(x)|<ε ................ (1)
Whenever 0<|x-a|< δ, this establishes the convergence of f(x). We now proceed to proving for g(x).
If ε>0 there exists a δ'>0 such that
|g(x)-L'|< ε.................... (2) whenever
0<|x-a|<δ'.
Suppose that 0<|x-a|<δ=min(δ,δ')
So that both equation (1) and equation                             (2) holds.
|(f+g)(x)-(L+L')|=|(f(x)-L)+(g(x)-L')|≤|f(x) -L|+|g(x)-L'|<2ε
Whivh proves for limf(x)+g(x)=L+L' as x→a.
(b)              In order to prove the scaler multiplication of the limit we also use the test of  convergence. cf(x)→cL

If ε>0 there exists a δ>0 such that |f(x)-L|<ε whenever |x-a|<δ, which proves f(x), now the scaler multiplication is |cf(x)-L|<ε where |c||f(x)-L|<ε since δ depends on the value of ε, then the limit exists, hence cf(x)→cL.
(c)          We now proceed to proving the composite function. Limf(x)g(x)=LL' as x→a, now we assume 0<|x—a|<δ whenever |f(x)g(x)-LL'|< ε
Now |f(x)g(x)—LL'|=|(fg)(x)—LL'|
=|(f(x)g(x)—L')+L'(f(x)—L)|
≤|f(x)||g(x)—L'|+|L'||f(x)—L|
≤(|f(x)|+|L'|)ε from equations (1) and (2).
≤(|f(x)—L|+|L|+|L'|)ε
≤(ε+|L|+|L'|)ε now from equation (1)
≤(1+|L|+|L'|)ε
Hence if ε<1 and x satisfies 0<|x—a|<δ, this proves limf(x)g(x)=LL' as x→a.
(d)           To Prove f(x)÷g(x)=L÷L' as x→a, we first observe that if L'≠0, there is δ''>0 such that |g(x)—L'|<|L'|/2
|g(x)|<|L'|/2........(3)
0<|x—a|<δ'', to see this, let L=L' and ε=|L'|/2, suppose that 0<|x—a|<min(δ,δ',δ''), so that equations (1),(2) and (3) holds.
                             Now









And hence the theorem is proved.