If #lim_(x rarr 2) (ax^2-b)/(x-2)=4# then find the value of a and b?
2 Answers
Explanation:
We have the following:
Whatever
Let's see if we can construct a difference of squares in the numerator, so we can cancel the
The difference of squares expression
If we have
Now we can evaluate this limit at
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!
# a= 1; b=4
Explanation:
We seek constants
# lim_(x rarr 2) (ax^2-b)/(x-2)=4 #
Noticing that the denominator is zero when
Thus we require that the numerator 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 #