Indeterminate Form of Type `0/0`

Related Calculator: Limit Calculator

We already talked about indeterminate forms of type `0/0`. We studied rational functions and performed algebraic manipulations to get rid of indetermination.

However, there are functions that are not rational but still habe indeterminate form of type `0/0`.

For example consider function `f(x)=(e^x-1)/x`. We want to analyze behaviour of this function near `x=0`. In other words we want to calculate `lim_(x->0)(e^x-1)/x`. Since `lim_(x->0)e^x-1=0` and `lim_(x->0)x=0` then we have indeterminate form of type `0/0`. But here we can't perform algebraic manipulations to simplify expression and get rid of determination. We just can't simplify it.

Luckily there is a powerful method that allows us to find such type of limit.

First 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)=0` and `lim_(x->a)g(x)=0`, 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.

First 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->0)(e^x-1)/x`.

Since `lim_(x->0)(e^x-1)=e^0-1=0` and `lim_(x->0)x=0` then we can apply L'Hopital's Rule:

`lim_(x->0)(e^x-1)/x=lim_(x->0)((e^x-1)')/(x')=lim_(x->0)(e^x)/1=e^0=1`.

Example 2. Find `lim_(x->0)(e^x-e^(-x))/(ln(e-x)+x-1)`.

Since both numerator and denominator approach 0 as `x->0` we can apply L'Hopital's Rule:

`lim_(x->0)(e^x-e^(-x))/(ln(e-x)+x-1)=lim_(x->0)((e^x+e^(-x))')/((ln(e-x)+x-1)')=lim_(x->0)(e^x+e^(-x))/(- 1/(e-x)+1)=(2)/(-1/e+1)=(2e)/(e-1)`.

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

Example 3. Find `lim_(x->0)(e^x-e^(-x)-2x)/(x-sin(x))`.

Since both numerator and denominator approach 0 as `x->0` we can apply L'Hopital's Rule:

`lim_(x->0)(e^x-e^(-x)-2x)/(x-sin(x))=lim_(x->0)((e^x-e^(-x)-2x)')/((x-sin(x))')=lim_(x->0)(e^x+e^(-x)-2)/(1-cos(x))`.

We again obtained indeterminate form of type `0/0` and apply L'Hopital's Rule once more:

`lim_(x->0)(e^x+e^(-x)-2)/(1-cos(x))=lim_(x->0)((e^x+e^(-x)-2)')/((1-cos(x))')=lim_(x->0)(e^x-e^(-x))/(sin(x))`.

We again obtained indeterminate form of type `0/0` so apply L'Hopital's Rule once more:

`lim_(x->0)(e^x-e^(-x))/(sin(x))=lim_(x->0)((e^x-e^(-x))')/((sin(x))')=lim_(x->0)(e^x+e^(-x))/(cos(x))=(e^0+e^(-0))/(cos(0))=2`.

So, `lim_(x->0)(e^x-e^(-x)-2x)/(x-sin(x))=2`.

Example 4. Find `lim_(x->0)(sin(x)-x)/(x^3)` .

Since `sin(x)-x->0` and `x^3->0` as `x->0` then

`lim_(x->0)(sin(x)-x)/x^3=lim_(x->0)((sin(x)-x)')/((x^3)')=lim_(x->0)(cos(x)-1)/(3x^2)` .

Since `cos(x)-1->0` and `3x^2->0` as `x->0` we again have indeterminate form of type `0/0` and apply L'Hopital's Rule once more:

`lim_(x->0)((cos(x)-1)')/((3x^2)')=lim_(x->0)(-sin(x))/(6x)`.

Again, we have indeterminate form of type `0/0` so apply L'Hopital's Rule third time:

`lim_(x->0)(-sin(x))/(6x)=lim_(x->0)((-sin(x))')/((6x)')=lim_(x->0)(-cos(x))/6=-cos(0)/6=-1/6`.

Example 5. Find `lim_(x->0^+)((sin(x))-1)/(cos(x))` .

If we blindly attempt to apply L'Hopital's Rule, we will get that `lim_(x->0^+)((sin(x)-1)')/((cos(x))')=lim_(x->0^+)(cos(x))/(-sin(x))=-oo`.

THIS IS WRONG! We can't apply L'Hopital's rule because `lim_(x->0^+)cos(x)=1` and we don't have indeterminate form.

In fact `lim_(x->0)(sin(x)-1)/cos(x)=(sin(0)-1)/(cos(0))=-1`.

Example 5 shows what can go wrong if you use L'Hopital's Rule without thinking checking conditions of theorem.

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 6. Find `lim_(x->2)(x^2-4)/(x-2)`.

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

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

Sometimes we need to combine L'Hopital's Rule with algebraic manipulations.

Example 7. Find `lim_(x->0)(tan(x)-x)/(x-sin(x))`.

We have indeterminate form of type `0/0` so can apply L'Hopital's rule:

`lim_(x->0)(tan(x)-x)/(x-sin(x))=lim_(x->0)((tan(x)-x)')/((x-sin(x))')=lim_(x->0)(1/(cos^2(x))-1)/(1-cos(x))`.

We again have indeterminate form of type `0/0` so we can apply L'Hopital's Rule once more. However, if we blindly use L'Hopital's Rule we of course will obtain correct answer, but we need to apply L'Hopital's Rule three times (you can check it) and taking derivatives won't be pleasant.

Instead, we perform some algebraic manipulations: `(1/(cos^2(x))-1)/(cos(x)-1)=(1-cos^2(x))/(cos^2(x)(1-cos(x)))=((1-cos(x))(1+cos(x)))/(cos^2(x)(1-cos(x)))=(1+cos(x))/(cos^2(x))`.

So, `lim_(x->0)(1/(cos^2(x))-1)/(1-cos(x))=lim_(x->0)(1+cos(x))/(cos^2(x))=(1+cos(0))/cos(0)=2`.

So, algebraic manipulations saved us time and efforts.

Example 8. Calculate `lim_(x->0)(xe^(2x)+xe^x-2e^(2x)+2e^x)/((e^x-1)^3)`.

Applying L'Hopital's Rule a couple of times and simplifying result between applications gives:

`lim_(x->0)(xe^(2x)+xe^x-2e^(2x)+2e^x)/((e^x-1)^3)=lim_(x->0)(e^(2x)+2xe^(2x)+e^x+xe^x-4e^(2x)+2e^x)/(3e^x(e^x-1)^2)=lim_(x->0)(2xe^x-3e^x+3+x)/(3(e^x-1)^2)=`

`=lim_(x->0)(2e^x+2xe^x-3e^x+1)/(6e^x(e^x-1))=lim_(x->0)(2x-1+e^(-x))/(6(e^x-1))=lim_(x->0)(2-e^(-x))/(6e^x)=1/6`.