Evaluate lim x-->0 sqrt(3+x) - sqrt(3) / x

To find the   limit of (sqrt(3+x)-sqrt3)/x  as
x-->0

Solution:

Substitition  x=
0 makes the expression 0/0 of indetermination.

(i)Using
differentiating technic we can evaluate the limit.

(ii) The
limit could be solved by rationalising the numerator  (by multiplying  numerator
(sqrt(3+x) -sqrt)  and denominator x  by the conjugate surd , i.e sqrt((
3+x)+sqrt3 ).

Using
diferentiating method:

We know that the definition of
differention is  given by:

f'(u) = Lt {f(u+x) - f(u)}/x as 
x-->0

Take f(u) = sqrt
u

Then {f'(u) at u= 3} = Lt {sqrt(3+x) -sqrt3}/x  as
x--> 0

But{ f'(u) at u =3}   = { (sqrt(u))' at u =3}
= {(1/2)u^(1/2 - 1) at 3 =3 }, as (u^n)' = nu^(n-1)

(f'(u)
at u = 3} =  (1/2)(3^(1/2 - 1) = 1/(2sqrt3).

Therefore Lt
(sqrt(3+x)-sqrt3)/x as x--> 0 = (sqrt u)' at u=3 = 1/(2sqrt3) . Ratinalise the
denominator.

Lt(sqrt(3+x)-sqrt3)/x = sqr3/(2sqrt3*sqrt3) =
(sqrt3)/6.

Second method:

Lt
(sqrt(3+x)-sqrt3)/x = Lt
(sqrt(3+x)-sqrt3)(sqrt(3+x)+sqrt3)/{xsqrt(3+x)+sqrt3)

Lt
(sqrt(3+x)-sqrt3)/x = Lt ((3+x)-3)/{xsqrt(3+x)+sqrt3)

Lt
(sqrt(3+x)-sqrt3)/x = Lt x/{xsqrt(3+x)+sqrt3). We reduce numaratot and denominator by
x.

Lt (sqrt(3+x)-sqrt3)/x = Lt
1/{sqrt(3+0)+sqrt3)

Lt (sqrt(3+x)-sqrt3)/x =  1/(2sqrt3) =
sqrt3/(2sqrt3*sqrt3) = sqrt3/6.