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

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!

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 #