Question

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

Answer 1

`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`

Answer 2

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`

Answer 3

`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`

Answer 4

`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.!