Evaluate lim_(x rarr -oo) sqrt(x^2 + x) - x ?

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1 Answer
May 3, 2018

lim_{x \to -\infty} [ sqrt(x^2 + x) - x ] = \infty

Explanation:

We know that

lim_{x \to -\infty} [ sqrt(x^2 + x) - x ] = lim_{x \to -\infty} sqrt(x^2 + x) + lim_{x \to -\infty} [ - x ] ,

and, trivially,

lim_{x \to -\infty} [- x] = \infty.

Therefore, the interesting part is the limit of the square root. If we look at the term under this square root, then, if x goes to -\infty, we have a positive part, x^2, and a negative part, x. However, as x^2 grows much faster than x decays, we know that the limit has to be \infty. We can prove this by showing that for an arbitrary \epsilon > 0 it holds that sqrt(x^2 + x) > \epsilon if x is chosen small enough (which is guaranteed to be the case at some point, as x goes to -\infty).

Starting the proof, we set

sqrt(x^2 + x) > \epsilon,

and thus get

x^2 + x > \epsilon^2.

By completing the square, we find that this is the same as

(x + 1/2)^2 - 1/4 > \epsilon^2,

and hence get

(x + 1/2)^2 > \epsilon^2 + 1/4

or

| x + 1/2 | > \sqrt(\epsilon^2 + 1/4).

Finally, noting that | x + 1/2 | >= | x | - 1 / 2, we arrive at

| x | > \sqrt(\epsilon^2 + 1/4) + 1/2 = {\sqrt{4 \epsilon^2 + 1} + 1} / 2.

This shows that

lim_{x \to -\infty} sqrt(x^2 + x) = \infty,

and it follows that

lim_{x \to -\infty} [ sqrt(x^2 + x) - x ] = \infty.