#limxrarra (x^2-a^2)/(sqrtx-sqrta)=32# What value should "a" have?

3 Answers
Narad T.
Sep 4, 2017

The value of #a=4#

Explanation:

To calculate the limit

#lim_(x->a)(x^2-a^2)/(sqrtx-sqrta)=(a^2-a^2)/(sqrta-sqrts)=0/0#

Which is an an indeterminate form

We apply l'Hôpital's rule

#lim_(x->a)(x^2-a^2)/(sqrtx-sqrta)=lim_(x->a)((x^2-a^2)')/((sqrtx-sqrta)')#

#=lim_(x->a)(2x)/(1/(2sqrtx))#

#=4asqrta#

Therefore,

#4asqrta=32#

#a^(3/2)=32/4=8#

#a=8^(2/3)#

#a=2^2=4#

#a=4#

Steve M
Sep 4, 2017

#a=4#

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 #

mason m
Sep 8, 2017

#a=4#

Explanation:

Factor as a difference of squares multiple times. You may want to recall that #u^2-v^2=(u+v)(u-v)#. Expand the right hand side to confirm this equality. Using this identity:

#lim_(xrarra)(x^2-a^2)/(sqrtx-sqrta)=lim_(xrarra)((x+a)(x-a))/(sqrtx-sqrta)#

We'll again use #u^2-v^2=(u+v)(u-v)#, but here #u=sqrtx# and #v=sqrta#.

#=lim_(xrarra)((x+a)((sqrtx)^2-(sqrta)^2))/(sqrtx-sqrta)#

#=lim_(xrarra)((x+a)(sqrtx+sqrta)(sqrtx-sqrta))/(sqrtx-sqrta)#

Now the #sqrtx-sqrta# terms cancel one another out:

#=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#

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