How do you find the limit #lim (2x+4)/(5-3x)# as #x->oo#?

Question

2447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T22:25:20

#= - 2/3#

#lim_(x to oo) (2x+4)/(5-3x)#

Divide numerator and deminitator by #x# :)

#= lim_(x to oo) (2+4/x)/(5/x-3)#

There are then various rules that allow you to do this, but it's also pretty intuitive:
#= (2+lim_(x to oo) 4/x)/(lim_(x to oo) 5/x-3)#

#= - 2/3#

Answer 2 · 2017-01-04T22:30:03

Given: #lim_(xrarroo) (2x + 4)/(5 -3x)#

Because the expression evaluated at the limit is the indeterminate form, #-oo/oo#, then one should use L'Hôpital's rule .

Compute the derivative of the numerator:

#(d(2x + 4))/dx = 2#

Compute the derivative of the denominator:

#(d(5 - 3x))/dx = -3#

Make a new expression:

#lim_(xrarroo) 2/-3#

The limit of the above expression is easy to evaluate:

#lim_(xrarroo) 2/-3 = -2/3#

The rule says that the limit of the original expression is the same:

#lim_(xrarroo) (2x + 4)/(5 -3x) = -2/3#