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

Question

Isaiah Allen · Accepted Answer

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

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#

Christopher Agresta · Answer

undefined To find the limit of (8x-14)/(sqrt(13x+49x^2)) as x approaches infinity, we can use the concept of limits.

First, we divide both the numerator and denominator by x^2, which gives us (8/x - 14/x)/(sqrt(13/x + 49)).

As x approaches infinity, 8/x and 14/x both approach 0. Additionally, 13/x and 49/x^2 both approach 0.

Therefore, the limit simplifies to (0 - 0)/(sqrt(0 + 0)) = 0/0.

To further evaluate this indeterminate form, we can apply L'Hôpital's Rule. Taking the derivative of the numerator and denominator, we get (8 - 14)/(1/2sqrt(13)) = -6/(1/2sqrt(13)) = -12/sqrt(13).

Hence, the limit of (8x-14)/(sqrt(13x+49x^2)) as x approaches infinity is -12/sqrt(13).