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How do you find the limit of #sqrt(x^2+2x)+x# as x approaches infinity?

Answer 1

The limit does not exist (it diverges to infinity)

We want # lim_(x rarr oo) sqrt(x^2+2x) + x #

Both #x^2# and #x# (and therefore #2x#) increase without bound as #x# increases, hence #sqrt(x^2+2x)# also increases without bound and the limit diverges.

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Answer 2

Paisley Johnson

To find the limit of sqrt(x^2+2x)+x as x approaches infinity, we can simplify the expression by dividing every term by x. This gives us sqrt(1+2/x)+1. As x approaches infinity, 2/x approaches 0, so the expression simplifies to sqrt(1+0)+1, which is equal to 2. Therefore, the limit of sqrt(x^2+2x)+x as x approaches infinity is 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.