What is the value of 1/(x - sqrt(x^2 + 2x)) for limit x equal to inf.

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