How do you find the limit of #(8x-14)/(sqrt(13x+49x^2))# as x approaches #oo#?

1 Answer
Ken C.
Jun 19, 2016

Do a little factoring and canceling to get #lim_(x->oo)(8x-14)/(sqrt(13x+49x^2))=8/7#.

Explanation:

At limits of infinity, the general strategy is to take advantage of the fact that #lim_(x->oo)1/x=0#. Normally that means factoring out an #x#, which is what we'll be doing here.

Begin by factoring an #x# out of the numerator and an #x^2# out of the denominator:
#(x(8-14/x))/(sqrt(x^2(13/x+49)))#
#=(x(8-14/x))/(sqrt(x^2)sqrt(13/x+49))#

The issue is now with #sqrt(x^2)#. It is equivalent to #abs(x)#, which is a piecewise function:
#abs(x)={(x, "for",x > 0),(-x,"for",x < 0):}#

Since this is a limit at positive infinity (#x>0#), we will replace #sqrt(x^2)# with #x#:
#=(x(8-14/x))/(xsqrt(13/x+49))#

Now we can cancel the #x#s:
#=(8-14/x)/(sqrt(13/x+49))#

And finally see what happens as #x# goes to #oo#:
#=(8-14/oo)/(sqrt(13/oo+49))#

Because #lim_(x->oo)1/x=0#, this is equal to:
#(8-0)/(sqrt(0+49))#
#=8/sqrt(49)#
#=8/7#

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