Limit value with an unknown

[Kebabsen123]

lim
x -> infinite = (sqrt(x^2 + x) - sqrt(x^2+ax)) = 7

Decide what a should be. What I've tried is to multiply with the conjugate both on
(sqrt(x^2 + x) - sqrt(x^2+ax)) and under. I've also tried to multiply both sides with the conjugate. I'm also thinking of dividing both sides with 7.
I'm really not sure what to do, can someone help me out? Thanks

 

[lev888]

Is there a denominator somewhere?

 

[Kebabsen123]

No

 

[pka]

lim
x -> infinite = (sqrt(x^2 + x) - sqrt(x^2+ax)) = 7
Decide what a should be. What I've tried is to multiply with the conjugate both on
(sqrt(x^2 + x) - sqrt(x^2+ax)) and under. I've also tried to multiply both sides with the conjugate. I'm also thinking of dividing both sides with 7.
I'm really not sure what to do, can someone help me out?

\(\displaystyle \mathop {\lim }\limits_{x \to \infty } \sqrt {{x^2} + x} - \sqrt {{x^2} + ax} = \mathop {\lim }\limits_{x \to \infty } \frac{{({x^2} + x) - (x^2 + ax)}}{{\sqrt {{x^2} + x} + \sqrt {{x^2} + ax} }}\)
Now what?

 

[pka]

\(\displaystyle \mathop {\lim }\limits_{x \to \infty } \sqrt {{x^2} + x} - \sqrt {{x^2} + ax} = \mathop {\lim }\limits_{x \to \infty } \frac{{({x^2} + x) - (x^2 + ax)}}{{\sqrt {{x^2} + x} + \sqrt {{x^2} + ax} }}\)

I think this needs to be finished for others to use.
\(\displaystyle \mathop {\lim }\limits_{x \to \infty } \sqrt {{x^2} + x} - \sqrt {{x^2} + ax}\\ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{({x^2} + x) - (x^2 + ax)}}{{\sqrt {{x^2} + x} + \sqrt {{x^2} + ax} }}\\ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{(1-a)x}}{{\sqrt {{x^2} + x} + \sqrt {{x^2} + ax} }}\\ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{(1-a)}}{{\sqrt {{1} + \frac{1}{x}} + \sqrt {{1} + \frac{a}{x}} }}\\=\dfrac{1-a}{2}\)
If one sets \(\displaystyle \dfrac{1-a}{2}=7\) then \(\displaystyle \bf{a=~?}\)

 

[HallsofIvy]

\(\displaystyle \frac{(1- a)x}{\sqrt{x^2+ x}+ \sqrt{x^2+ ax}}\)

Once you got to this point, I would argue that, for very large x, both "x" and "ax" are very small compared to "\(\displaystyle x^2\)" and can be ignored so this is equivalent to \(\displaystyle \frac{(1- a)x}{\sqrt{x^2}+ \sqrt{x^2}}= \frac{(1- a)x}{2x}= \frac{1- a}{2}\).

Of course, what pka did is more "formal".