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

##### 1 Answer

# 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

# 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

# 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

# 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.