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

First, we'll check to see if we have an indetermination case. For this reason, we'll substitute x by 0.

lim [sqrt(3+x) - sqrt(3)] / x = [sqrt(3+0) - sqrt3]/0 = 0/0

"0/0" is an indetermination, so we could use l'Hospital rule.

If lim (f/g) = 0/0, then lim (f/g)= lim (f')/(g')

lim [sqrt(3+x) - sqrt(3)] / x = lim [sqrt(3+x) - sqrt(3)]' / (x)'

[sqrt(3+x) - sqrt(3)]' = (3+x)'/2sqrt(3+x) - 0

where (3+x)' = 0 + 1 = 1

[sqrt(3+x) - sqrt(3)]' = 1/2sqrt(3+x)

x' = 1

lim [sqrt(3+x) - sqrt(3)] / x = lim 1/1*2sqrt(3+x)

Now, we'll substitute x by 0:

lim 1/1*2sqrt(3+x) = 1/2*sqrt(3+0)

lim 1/1*2sqrt(3+x) = 1/2sqrt3

lim 1/1*2sqrt(3+x) = sqrt3/2*3

lim [sqrt(3+x) - sqrt(3)] / x = sqrt3/6

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.