Question

Limit of (ax^2-b)/(x-2)=4 as x approaches to 2 find the value of a and b?

Answer 1

`a=1 \ \` and `\ \ b=4`

Explanation:

We require that:

`L = lim_(x rarr 2) (ax^2-b)/(x-2)=4`

As the denominator is zero when `x=2` then the limit can only exist if the numerator has a factor `(x-2)`, Then by the remainder theorem, the numerator has a root `x=2`, let us denote the other root by `alpha` then we have:

`L = lim_(x rarr 2) ((x-alpha)(x-2))/(x-2)`

`\ \ = lim_(x rarr 2) (x-alpha)`

`\ \ = 2-alpha`

But we know that `L=4`, and so:

`2-alpha = 4 => alpha=-2`

Given this we can now write the numerator as:

`ax^2-b -= (x+2)(x-2) = x^2-4`

And by comparing coefficients we have:

`a=1 \ \` and `\ \ b=4`