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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