How do you evaluate the limit #(6x+1)/(2x+5)# as x approaches #oo#?

Question

Elijah Abram · Accepted Answer

#lim_(x to oo) (6x+1)/(2x+5)# #= lim_(x to oo) (6+1/x)/(2+5/x)# #= (6+ lim_(x to oo)  1/x)/(2+ lim_(x to oo)  5/x)# #= 6/2 = 3#

Cameron Alexander · Answer

Because the expression evaluated at the limit results in an indeterminate form #oo/oo#, one should use L'Hôpital's rule .

L'Hôpital's rule states that, if you take the derivative of the numerator and the derivative of the denominator, the resulting fraction goes to the same limit as the original. #lim_(xrarroo) (6x+1)/(2x+5) =# #lim_(xrarroo) ((d(6x+1))/dx)/((d(2x+5))/dx) =# #lim_(xrarroo) 6/2 = 3# Therefore, the limit of the original expression is: #lim_(xrarroo) (6x+1)/(2x+5) =3#

Charles Anctil · Answer

#3# #"divide terms on numerator/denominator by x"# #((6x)/x+1/x)/((2x)/x+5/x)=(6+1/x)/(2+5/x)# #rArrlim_(xtooo)(6x+1)/(2x+5)# #=lim_(xtooo)(6+1/x)/(2+5/x)# #=(6+0)/(2+0)=3#

Cameron Alexander · Answer

#3.# Let #x=1/y," so that, as "x to oo, y to 0.# #"The Limit="lim_(y to 0) (6/y+1)/(2/y+5),# #=lim_(y to 0) (6+y)/(2+5y)=(6+0)/(2+5(0))=6/2,# #:." The Limit="3.# Enjoy Maths.!

Paisley Johnson · Answer

undefined To evaluate the limit of (6x+1)/(2x+5) as x approaches infinity, we can use the concept of limits at infinity. By dividing both the numerator and denominator by x, we get (6+1/x)/(2+5/x). As x approaches infinity, 1/x approaches 0. Therefore, the limit simplifies to 6/2, which equals 3. Hence, the limit of (6x+1)/(2x+5) as x approaches infinity is 3.