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:

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