Question

If `lim_(x rarr 2) (ax^2-b)/(x-2)=4` then find the value of a and b?

Answer 1

`a=1`, `b=4`

Explanation:

We have the following:

`lim_(xto2)(color(blue)((ax^2-b))/(x-2))`

Whatever `a` and `b` happen to be, if we plug `2` into the expression the way it is now, we get a zero in the denominator, which makes the expression undefined.

Let's see if we can construct a difference of squares in the numerator, so we can cancel the `x-2` in the bottom.

The difference of squares expression `color(blue)(x^2-4)` can be factored as `color(blue)((x+2)(x-2))`, which allows us to cancel `x-2` on the bottom.

If we have `x^2-4` in the numerator, this means `a=1` and `b=4`. We now have

`lim_(xto2)((x^2-4)/(x-2))=lim_(xto2)(((x+2)cancel(x-2))/cancel(x-2))`

`lim_(xto2)(x+2)=4`

Now we can evaluate this limit at `x=2`, and we indeed get `4`.

Notice, if we evaluated the original limit, we would get an undefined expression, so my next thought was to cancel the denominator.

Hope this helps!

Answer 2

# a= 1; b=4

Explanation:

We seek constants `a,b in RR` such that:

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

Noticing that the denominator is zero when `x=2`, and that the limit exists it must be that the limit is of an indeterminate form `0/0`, as otherwise the limit could not exist.

Thus we require that the numerator is `0` when `x=2`, that is

`[ax^2-b]_(x=2) = 0`
`\ \ => 4a-b = 0`
`\ \ => b=4a` ..... [A]

If we apply L'Hôpital's rule then we know that for an indeterminate form, then

`lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x))`

So differentiating the numerator and denominator independently, then we have:

`lim_(x rarr 2) (d/dx(ax^2-b))/(d/dx(x-2))=4`
`\ \ => lim_(x rarr 2) (2ax)/(1)=4`
`\ \ => 4a=4`
`\ \ => a=1`

And using [A], we have:

`b = 4 xx 1= 4`