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

4 Answers
Eddie
Apr 16, 2017

#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#

Douglas K.
Apr 16, 2017

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

Explanation:

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#

Jim G.
Apr 16, 2017

#3#

Explanation:

#"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#

Ratnaker Mehta
Apr 16, 2017

#3.#

Explanation:

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.!

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