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# Whats the limit of?

Lim square root of ( (x^2+ax) - square root of(x^2+bx) )

x increase without bound (+ infinity)

### 3 Answers

- Mark PLv 51 decade agoFavorite Answer
The limit is infinity.

Reasoning:

As x approaches infinity, you can throw out the ax and bx terms because x^2 dominates.

So sqrt(x^2-sqrt(x^2)) = sqrt(x^2 - x)

Again, the x^2 dominates, so you simply have:

lim(x->infinity) of x

which is simply infinity.

Now, if you misplaced your parentheses in the problem statement, the answer is more interesting.

sqrt(x^2+ax)-sqrt(x^2+bx)

lim(x->inf) of sqrt(x^2+ax) is x+a/2, since (x+a/2)^2 = (x^2+ax+1)

lim(x->inf) of sqrt(x^2+bx) is x+b/2

So lim(x->inf) of the difference is (x+a/2)-(x+b/2) or (a-b)/2.

This is confirmed with calculations using very large x and small a and b.

- Phineas BoggLv 61 decade ago
I believe it is (a-b)/2.

My reasoning:

x^2 + ax + (a/2)^2 = (x+a/2)^2 and

x^2 + bx + (b/2)^2 = (x+b/2)^2

So the limit looks like limit:

sqrt[(x+a/2)^2] - sqrt[(x+b/2)^2] =

x+a/2 - (x+b/2) = a/2 - b/2

It's not formal, but if I had to put money on this problem, I would put it on (a-b)/2