Question

If `lim_(x rarr oo)((ax^2+bx+3)^(1/2) -2x)=7`, then find `a` and `b`?

Answer 1

`lim_{x rightarrow infty} (sqrt{4x^2+28x+3} - 2x) = 7`

Explanation:

Complete the square underneath the radical to get:

`lim_{x rightarrow infty}( sqrt{a}sqrt{(x+(b)/(2a))^2 - {b^2 }/{4ac} + 3/a} -2x )`
`= lim_{x rightarrow infty} (xsqrt(a) + b/(2sqrt(a)) -2x )= 7`

Okay, so we have this equivalent form, and note that this is the equation of the line, and so for it to have a finite limit of seven, the `x` terms above have to go to 0.

Thus `xsqrta -2x = 0 Rightarrow (x)(sqrta -2) = 0 Rightarrow sqrta =2 Rightarrow a = 4`

In this case our limit becomes

`lim_{x rightarrow infty} (xsqrt(a) + b/(2sqrt(a)) -2x) = lim_{x rightarrow infty} (xsqrt(4) + b/(2sqrt(4)) -2x ) = b/4 = 7 Rightarrow b = 28`

So in fact this is satisfied only by `a=4` and `b=28`