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

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Answer 1 · 2017-12-12T08:39:57

#{(a=1),(b=4):}#

We have:

#(1)# #lim_(x->2) (ax^2-b)/(x-2) = 4#

As the denominator is null in #x=2#, for the limit to be finite also the numerator must be null, so:

#[ax^2-b]_(x=2)=0#

#4a=b#

and #(1)# becomes:

#lim_(x->2) (ax^2-4a )/(x-2) = 4#

Now:

#(ax^2-4a )/(x-2) = (a (x^2-4))/(x-2) = (a(x+2)(x-2))/(x-2) = a(x+2)#

So:

#lim_(x->2) a(x+2) = 4#

which means #a=1# and in conclusion:

#{(a=1),(b=4):}#