Here if replace x with 0 we get the result as 0/0 which is indeterminate. Therefore we can use L’ Hopital’s Rule which states that if an expression of the form lim x-->0 [f(x)/ g(x)] gives the indeterminate form 0/0, the limit can be found as [f’(x)/g’(x)] for x=0.

Now for the given expression [(sqrt (1+x) – 1 –(x/2)) / x^2], f(x) = (sqrt (1+x) – 1 –(x/2)) and g(x) = x^2

f’(x) = (1/2) (1+ x) ^ (-1/2) – (1/2)

g’(x) = 2x

Now we have [f’(x)/g’(x)] for x=0 as ((1/2) (1+ x) ^ (-1/2) – (1/2)) / 2x = 0/0 again.

So we take the differential again

f’’(x) = - (1/4) (1+x) ^ (-3/2)

g’’(x) = 2

If we determine [-(1/4) (1+x) ^ (-3/2) / 2] for x =0= -1/8

**Therefore the result is -1/8.**

First, we'll substitute x by the given value 0, in the expression of the limit:

lim [(sqrt (1+x) – 1 –(x/2)) / x^2] = [sqrt(1+0) - 1 - 0/2]/0

lim [(sqrt (1+x) – 1 –(x/2)) / x^2] = (1-1)/0

lim [(sqrt (1+x) – 1 –(x/2)) / x^2] = 0/0

We've obtained an indeterminacy case 0/0.

We'll apply L'Hospital rule:

lim f'(x)/g'(x) = lim f(x)/g(x)

f'(x) = [sqrt (1+x) – 1 –(x/2)]'

f'(x) = (1+x)'/2sqrt(1+x) - 0 - 1/2

f'(x) = 1/2sqrt(1+x)- 1/2

g'(x) = 2x

lim [(sqrt (1+x) – 1 –(x/2)) / x^2] = lim [1/2sqrt(1+x)- 1/2]/2x

We'll substitute x by 0:

lim [1/2sqrt(1+x)- 1/2]/2x = [1/2sqrt(1+0)- 1/2]/2*0 = 0/0

Since we've obtained again an indeterminacy, we'll differentiate the result:

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

f"(x) = [-2/2sqrt(1+x)]/4(1+x) - 0

f"(x) = [-1/sqrt(1+x)]/4(1+x)

g"(x) = 2

lim [1/2sqrt(1+x)- 1/2]/2x = lim -1/8(1+x)sqrt(1+x)

We'll substitute x by 0:

lim -1/8(1+x)sqrt(1+x) = -(1/8)*(1/1)