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How do you find #lim sqrt(x^2+x)-x# as #x->oo#?

Answer 1

Isabella Ackerman

I found #1/2#

I would try a kind of reverse rationalization:

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Answer 2

Charles Arocha

To find the limit of sqrt(x^2+x)-x as x approaches infinity, we can simplify the expression. By multiplying the numerator and denominator by the conjugate of the expression, we can eliminate the square root:

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

Now, we can divide both the numerator and denominator by x:

lim x / (sqrt(x^2+x)+x) as x->oo = lim 1 / (sqrt(1+1/x)+1) as x->oo

As x approaches infinity, 1/x approaches 0. Therefore, we can simplify further:

lim 1 / (sqrt(1+1/x)+1) as x->oo = 1 / (sqrt(1+0)+1) = 1 / (sqrt(1)+1) = 1 / (1+1) = 1 / 2

Hence, the limit of sqrt(x^2+x)-x as x approaches infinity is 1/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find #lim sqrt(x^2+x)-x# as #x-&gt;oo#?

How do you find #lim sqrt(x^2+x)-x# as #x->oo#?