Question

How do you find the limit of `(x+sin x ) / ( 3x + cosx )` as x approaches infinity using l'hospital's rule?

Answer 1

The limit `1/3`

Explanation:

L'Hospital's rule states the that:

if `lim_(x->h) f(x)/g(x) -> 0/0` then`lim_(x->h) f(x)/g(x)=lim_(x->h) (f'(x))/(g'(x))`

The same also applies for indeterminate form of `oo/oo` which is the case we have here.

So we can differentiate the top of the fraction and the bottom of the fraction and then try to evaluate the limit again.

So:

`d/dx(x+sin(x)) = 1 + cos(x)`
and
`d/dx(3x+cos(x)) = 3-sin(x)`

So from l'hospital's rule we can say that:

`lim_(x->oo) (x+sin(x))/(3x+cos(x))=lim_(x->oo) (1+cos(x))/(3-sin(x))`

We can say that`lim_(x->oo)cos(x)=0` and `lim_(x->oo)sin(x) =0` as these functions just oscillate around 0 infinitely so:

`lim_(x->oo) (1+cos(x))/(3-sin(x)) =1/3`

Doing a graph of the function we can see that the graph oscillates around `1/3`.

graph{(x+sin(x))/(3x+cos(x)) [-0.82, 9.18, -2.66, 2.34]}