We have to find lim x--> inf, [1/(x - sqrt(x^2 + 2x))].

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

=> (x + sqrt(x^2 + 2x)/ (x - sqrt(x^2 + 2x)* (x + sqrt(x^2 + 2x)

=> (x + sqrt(x^2 + 2x)/ x^2 - x^2 - 2x

=> (x + sqrt(x^2 + 2x)/-2x

=> (x/-2x) – sqrt (x^2/4x^2 + 2x/4x^2)

=> -1/2 – sqrt (1/4 + 1/2x)

The required limit is now:

lim x--> +inf [-1/2 – sqrt (1/4 + 1/2x)]

substitute x = + inf, 1/x = 0

=> -1/2 – sqrt (1/4)

=> -1/2 – 1/2

=> -1

**The required result of the limit is -1.**

To determine the limit of the fraction, we'll substitute infinite into the expression and we'll get, at denominator, the indeterminacy case: infinite - infinite.

We'll multiply the fraction by the conjugate of denominator, that is x + sqrt(x^2 + 2x).

We'll get to the denominator the difference of squares:

[x - sqrt(x^2 + 2x)][x + sqrt(x^2 + 2x)] = x^2 - x^2 - 2x

We'll eliminate like terms and we'll get:

[x - sqrt(x^2 + 2x)][x + sqrt(x^2 + 2x)] = -2x

We'll re-write the fraction:

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

We'll apply limit both sides:

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

We'll factorize by x the numerator:

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

We'll simplify and we'll get:

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

**lim 1/[x - sqrt(x^2 + 2x)] = -1, x -> infinite**