#limxrarra (x^2-a^2)/(sqrtx-sqrta)=32# What value should "a" have?
3 Answers
The value of
Explanation:
To calculate the limit
Which is an an indeterminate form
We apply l'Hôpital's rule
Therefore,
Explanation:
We have:
# L = lim_(x rarr a) (x^2-a^2)/(sqrtx-sqrta) = 32 #
Which we can write as:
# L = lim_(x rarr a) (x^2-a^2)/(sqrtx-sqrta) * (sqrtx+sqrta)/(sqrtx+sqrta)#
# \ \ \ = lim_(x rarr a) ((x+a)(x-a))/(x-a) * (sqrtx+sqrta)#
# \ \ \ = lim_(x rarr a) (x+a) (sqrtx+sqrta)#
# \ \ \ = (a+a) (sqrta+sqrta)#
# \ \ \ = (2a) (2sqrta)#
# \ \ \ = 4a^(3/2)#
So if:
# L = 32 => 4a^(3/2) =32 #
# :. a^(3/2) =8 #
# :. a \ \ = 4 #
Explanation:
Factor as a difference of squares multiple times. You may want to recall that
#lim_(xrarra)(x^2-a^2)/(sqrtx-sqrta)=lim_(xrarra)((x+a)(x-a))/(sqrtx-sqrta)#
We'll again use
#=lim_(xrarra)((x+a)((sqrtx)^2-(sqrta)^2))/(sqrtx-sqrta)#
#=lim_(xrarra)((x+a)(sqrtx+sqrta)(sqrtx-sqrta))/(sqrtx-sqrta)#
Now the
#=lim_(xrarra)(x+a)(sqrtx+sqrta)#
#=lim_(xrarra)(a+a)(sqrta+sqrta)#
#=(2a)(2sqrta)#
#=4a^(3/2)=32#
Thus:
#a^(3/2)=8#
#a=(2^3)^(2/3)=2^2=4#