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