# 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'HÃ´pital'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 #