# Indeterminate Forms of Type oo/oo

## Related Calculator: Limit Calculator

Similarly there are limit of functions that represent indeterminate form of type oo/oo, but can't be calculated using algebraic manipulations.

However, there is corresponding L'Hopital's Rule that allows to handle indeterminate form of type oo/oo.

Second L'Hopital’s Rule. Suppose f(x) and g(x) are differentiable on (a,b] and g'(x)!=0 on (a,b]. If lim_(x->a)f(x)=oo and lim_(x->a)g(x)=oo, then lim_(x->a)(f(x))/(g(x))=lim_(x->a)(f'(x))/(g'(x)) if the limit on the right side exists (or is oo or -oo ).

It is especially important to verify the conditions before using L'Hopital's Rule.

Second L'Hopital’s Rule is also valid for one-sided limits and for limits at infinity or negative infinity; that is, "x->a" can be replaced by any of the following symbols: x->a^+, x->a^-, x->oo , x->-oo.

Example 1. Find lim_(x->oo)(ln(x))/(sqrt(x)).

Since lim_(x->oo)(ln(x))=oo and lim_(x->oo)sqrt(x)=oo then we can apply L'Hopital's Rule:

lim_(x->oo)(ln(x))/(sqrt(x))=lim_(x->oo)((ln(x))')/((sqrt(x))')=lim_(x->oo)(1/x)/(1/(2sqrt(x)))=lim_(x->oo)2/(sqrt(x))=0.

Sometimes we need to apply L'Hopital's rule more than once.

Example 2. Find lim_(x->oo)(e^x)/(x^2).

Since lim_(x->oo)e^x=oo and lim_(x->oo)x^2=oo then we can use L'Hopital's Rule:

lim_(x->oo)(e^x)/(x^2)=lim_(x->oo)((e^x)')/((x^2)')=lim_(x->oo)(e^x)/(2x).

Since e^x->oo and 2x->oo as x->oo then we still have indeterminate form of type (oo)/(oo) and we apply L'Hopital's rule once more:

lim_(x->oo)(e^x)/(2x)=lim_(x->oo)((e^x)')/((2x)')=lim_(x->oo)(e^x)/2=oo.

Example 3. Find lim_(x->oo)(1+1/x)/(x+1) .

If we blindly attempt to apply L'Hopital's Rule, we will get that lim_(x->oo)((1+1/x)')/((x+1)')=lim_(x->oo)(1-1/x^2)/(1)=1.

THIS IS WRONG! We can't apply L'Hopital's rule because lim_(x->oo)(1+1/x)=1 and we don't have indeterminate form.

In fact lim_(x->oo)(1+1/x)/(x+1)=lim_(x->oo)((x+1)/x)/(x+1)=lim_(x->oo)1/x=0.

Example 3 shows what can go wrong if you use L'Hopital's Rule without thinking checking conditions of theorem.

Now let's see what will be if we ignore condition that limit of ratio of derivatives should exist.

Example 4. Caclulate lim_(x->oo)(x+sin(x))/x.

We have indeterminate form here, so can apply L'Hopital's Rule:

lim_(x->oo)((x+sin(x))')/(x')=lim_(x->oo)(1+cos(x))/1.

Since cos(x) oscillates infinitely many times as x->oo then lim_(x->oo)cos(x) doesn't exist. Therefore lim_(x->oo)(1+cos(x)) doesn't exist.

However, initial limit exist: lim_(x->oo)(x+sin(x))/x=lim_(x->oo)(1+sin(x)/x)=1.

So, we need to be sure that limit of ratio of derivative exists, otherwise L'Hopital's Rule is inapplicable.

Other limits can be found using L'Hopital's Rule but are more easily found by other methods. So when evaluating any limit, you should consider other methods before using L'Hopital's Rule.

Example 5. Find lim_(x->oo)(x^2-4)/(2x^2-2).

Applying L'Hopital's Rule gives lim_(x->oo)(x^2-4)/(2x^2-2)=lim_(x->oo)((x^2-4)')/((2x^2-2)')=lim_(x->oo)(2x)/(4x)=lim_(x->oo)1/2=1/2.

But it is more natural to use algebraic manipulations: lim_(x->oo)(x^2(1-4/x^2))/(x^2(2-2/x^2))=lim_(x->oo)(1-4/x^2)/(2-2/x^2)=1/2.