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

1 Answer
Jan 25, 2018

# 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 #