Question

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

Answer 1

`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.