Limit of function with roots

[minor imp]

Not sure how to go about solving this

[crude lark]

Rationalize

@minor imp

[minor imp]

Using the conjugate correct?

I got that much but got stuck after that

[crude lark]

What did you get

$\frac{x^2+ax-x^2-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$

[little tulip]

Ephesians 2:8-9

[crude lark]

@minor imp

You're stuck there?

You cancel out the squared x's in the numerator, then divide all by highest exponent x which is just x

Which becomes x^2 in the square roots

And viola

[minor imp]

okay the divide by the highest exponent x

it's not sqrt(x)?

cause ax and bx are under the roots too

so wouldn't the x's exponent be 1/2 or am i doing that wrong

[onyx chasm]

The highest exponent. 1 > 1/2.

[minor imp]

huh

[onyx chasm]

...is that a question?

[crude lark]

Yes, you're ignoring the ax and bx in the numetator, whose exponent of x is 1 and higher than sqrt(x)

You're also ignoring the sqrt(x^2), which is also greater

[crude lark]

$\frac{ax-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$

[little tulip]

Ephesians 2:8-9

[crude lark]

$\frac{\frac{ax}{x}-\frac{bx}{x}}{\sqrt{\frac{x^2}{x^2}+\frac{ax}{x^2}}+\sqrt{\frac{x^2}{x^2}+\frac{bx}{x^2}}}$

[little tulip]

Ephesians 2:8-9

[analog vine]

Alternatively, you can do this. I also assume that by ∞ we mean +∞.
√(x^2 + ax) - √(x^2 + bx) = |x|(√(1 + a/x) - √(1 + b/x)) = (√(1 + a/x) - √(1 + b/x))/(1/x), x -> +∞
Let x = 1/t, t -> 0. Then we get:
(√(1 + at) - √(1 + bt))/t, t -> 0
So, now let f(x) = √(1 + x). We have:
(f(1 + at) - f(1 + bt))/t = (a - b)(f(1 + at) - f(1 + bt))/(at - bt), t -> 0
Apart from the first factor (a - b), the rest is just the definition of the derivative. So:
(a - b)(f(1 + at) - f(1 + bt))/(at - bt) -> (a - b)f'(0), t -> 0
Now, let's find f'(0).
f'(x) = 1/(2√(1 + x))
f'(0) = 1/2
So, the limit is (a - b)/2.

[onyx chasm]

Unsigned always defaults to positive. Unsigned radicals, unsigned infinity.