How do you find #lim (5+x^-1)/(1+2x^-1)# as #x->oo# using l'Hospital's Rule or otherwise?

1 Answer
Jan 10, 2017

# lim_( x rarr oo) (5+x^-1)/(1+2x^-1) =5#

Explanation:

In its present form it should be clear that you do not need to apply L'Hôpital's rule as #x^-1 rarr 0# as #x rarr oo# and so;

# lim_( x rarr oo) (5+x^-1)/(1+2x^-1) = (5+0)/(1+0) =5#

If you do want to apply L'Hôpital's rule the we should multiply numerator and denominator both by #x# to get;

# lim_( x rarr oo) (5+x^-1)/(1+2x^-1) = lim_( x rarr oo) x/x(5+x^-1)/(1+2x^-1)#
# " " = lim_( x rarr oo) (5x+1)/(x+2)#

We now have an indeterminate form of the type #oo/oo# so we can apply L'Hôpital's rule:

# lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x)) #

having satisfied ourselves that our limit meets L'Hôpital's criteria to get

# lim_( x rarr oo) (5+x^-1)/(1+2x^-1) = lim_( x rarr oo) (d/dx(5x+1))/(d/dx(x+2))#
# " " = lim_( x rarr oo) 5/1 = 5#, as above.