What is the limit: lim x-->0 [(sqrt (1+x) – 1 –(x/2)) / x^2]

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)

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