# Other Indeterminate Forms

## Related Calculator: Limit Calculator

We already talked about other indeterminate forms (indeterminate differences, indeterminate products and indeterminate powers), so we know that we can convert them into either indeterminate form of type `0/0` or indeterminate form of type `oo/oo`. This allows us to use either First or Second L'Hopital's Rules.

**Indeterminate Products**

Product will have indeterminate form only if `lim_(x->a)f(x)=0` and `lim_(x->a)g(x)=oo` (or `-oo` ). In this case `lim_(x->a)f(x)g(x)` is indeterminate form of type `0*oo`.

We can deal with it by writing product as a quotient:

`lim_(x->a)f(x)/(1/g(x))` (this will give indeterminate form of type `0/0`) or `lim_(x->a)g(x)/(1/f(x))` (this will give indeterminate form of type `(oo)/(oo)`).

**Example 1**. Find `lim_(x->0^+)xln(x)`.

Since `x->0` and `ln(x)->-oo` when `x->0^+` this is indeterminate form of type `0*oo`.

Representing product as quotient we obtain indeterminate form of type `(oo)/(oo)` and therefore can apply Second L'Hospital's Rule:

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

`=lim_(x->0^+)-x=0`.

Note, that we could represent product as indeterminate form of type `0/0`: `lim_(x->0^+)xln(x)=lim_(x->0^+)x/(1/(ln(x)))` . But applying L'Hopital's Rule to this limit will give a more complicated expression than the one we started with.

In general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit.

**Indeterminate Differences**

Difference will have indeterminate form only if `lim_(x->a)f(x)=oo` and `lim_(x->a)g(x)=oo`. In this case limit `lim_(x->a)(f(x)-g(x))` is indeterminate form of type `oo-oo`.

We can deal with it by rewriting it as quotient:

`f(x)-g(x)=1/(1/(f(x)))-1/(1/(g(x)))=(1/(g(x))-1/(f(x)))/(1/(f(x))*1/(g(x)))`. However, on practice there is a much simpler expression.

**Example 2**. Compute `lim_(x->(pi/2)^-)(sec(x)-tan(x))`.

Note that `sec(x)->oo` and `tan(x)->oo` as `x->(pi/2)^-`. Therefore we have indeterminate for of type `oo-oo` .

Here we can use a common denominator:

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

We can use L'Hopital's rule now, because `lim_(x->(pi/2)^-)(1-sin(x))=0` and `lim_(x->(pi/2)^-)cos(x)=0`:

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

**Example 3**. Calculate `lim_(x->0)(cot^2(x)-1/x^2)`.

Clearly we have indeterminate form of type `oo-oo` here.

Let's simplify this expression: `cot^2(x)-1/x^2=(cos^2(x))/(sin^2(x))-1/x^2=(x^2cos^2(x)-sin^2(x))/(x^2sin^2(x))=((xcos(x)+sin(x))(xcos(x)-sin(x)))/(x^2sin^2(x))=`

`=(xcos(x)+sin(x))/(x)*(xcos(x)-sin(x))/(xsin^2(x))`.

We can found limit of first factor very easy: `lim_(x->0)(xcos(x)+sin(x))/x=lim_(x->0)(cos(x)+(sin(x))/x)=cos(0)+1=2`.

Limit of second factor is indeterminate form of type `0/0` so we can apply L'Hopital's Rule:

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

`=lim_(x->0)(-x)/(sin(x)+2xcos(x))=lim_(x->0)(-1)/((sin(x))/x+2cos(x))=(-1)/(1+2*cos(0))=-1/3`.

Therefore,

`lim_(x->0)(cot^2(x)-1/x^2)=lim_(x->0)((xcos(x)+sin(x))/(x)*(xcos(x)-sin(x))/(xsin^2(x)))=`

`=lim_(x->0)(xcos(x)-sin(x))/(x)*lim_(x->0)(xcos(x)+sin(x))/(xsin^2(x))=2*(-1/3)=-2/3`.

**Indeterminate Powers**

Several indeterminate forms arise from the `lim_(x->a)(f(x))^(g(x))` .

- `lim_(x->a)f(x)=0` and `lim_(x->a)g(x)=0` give indeterminate form of type `0^0` (it is indeterminate because `0^x=0` for any `x>0` but `x^0=1` for any `x!=0` ).
- `lim_(x->a)f(x)=oo` and `lim_(x->a)g(x)=0` give indeterminate form of type `oo^0`.
- `lim_(x->a)f(x)=1` and `lim_(x->a)g(x)=+-oo` give indeterminate form of type `1^(oo)`.

Each of these three cases can be treated either by:

- Taking the natural logarithm: let `y=(f(x))^(g(x))` then `ln(y)=ln((f(x))^(g(x)))=g(x)ln(f(x))`.
- Writing the function as an exponential: `(f(x))^(g(x))=e^(g(x)ln(f(x)))`.

In either method we are led to the indeterminate product `g(x)ln(f(x))` , which is of type `0*oo`.

**Example 4**. Find `lim_(x->0^+)(1+sin(2x))^(cot(x))` .

First, make sure that this is indeterminate limit: since `1+sin(2x)->1` and `cot(x)->oo` as `x->0^+` then this is indeterminate form of type `1^(oo)`.

Let `y=(1+sin(2x))^(cot(x))` then `ln(y)=cot(x)ln(1+sin(2x))=(ln(1+sin(2x)))/(tan(x))`.

Since `ln(1+sin(2x))->0` and `tan(x)->0` as `x->0^+` , we have indeterminate form of type `0/0` and can apply L'Hopital's Rule:

`lim_(x->0^+)ln(y)=lim_(x->0^+)(ln(1+sin(2x)))/(tan(x))=lim_(x->0^+)((ln(1+sin(2x)))')/((tan(x))')=`

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

We found limit of `ln(y)`, but we need limit of `y`, so

`lim_(x->0^+)(1+sin(2x))^(cot(x))=lim_(x->0^+)y=lim_(x->0^+)e^(ln(y))=e^(lim_(x->0^+)ln(y))=e^2`.

**Example 5**. Calculate `lim_(x->0^+)x^x`.

Notice that this is indeterminate form of type `0^0` . We can proceed as in Example 4, but let's do it in another way:

`lim_(x->0^+)x^x=lim_(x->0^+)e^(ln(x^x))=lim_(x->0^+)e^(xln(x))=e^(lim_(x->0^+)xln(x))=e^0=1` because `lim_(x->0^+)xln(x)=0`.

**Example 6**. Find `lim_(x->0)(sin(x)/x)^(1/(1-cos(x)))`.

Here we have indeterminate form of type `1^oo`.

Let `y=(sin(x)/x)^(1/(1-cos(x)))`. We can consider the case `x>0` (because `y` is odd function).

So, we have that `ln(y)=ln((sin(x)/x)^(1/(1-cos(x))))=1/(1-cos(x))ln(sin(x)/x)=(ln(sin(x))-ln(x))/(1-cos(x))`.

Now, applying of L'Hopital's Rule gives the following:

`lim_(x->0)ln(y)=lim_(x->0)(ln(sin(x))-ln(x))/(1-cos(x))=lim_(x->0)((ln(sin(x))-ln(x))')/((1-cos(x))')=lim_(x->0)((cos(x))/(sin(x))-1/x)/(sin(x))=`

`=lim_(x->0)(xcos(x)-sin(x))/(xsin^2(x))=-1/3` (as was found in Example 3).

So, `lim_(x->0)y=e^(-1/3)=1/(root(3)(e))`.

**Example 7**. Find `lim_(x->oo)(pi/2-arctan(x))^(1/ln(x))`.

Since `pi/2-arctan(x)->0` and `1/(ln(x))->0` as `x->oo`, we have indeterminate form of type `0^0`.

Let `y=(pi/2-arctan(x))^(1/(ln(x)))`, then `ln(y)=ln((pi/2-arctan(x))^(1/(ln(x))))=1/(ln(x))ln(pi/2-arctan(x))`.

So, `lim_(x->oo)ln(y)=lim_(x->oo)(ln(pi/2-arctan(x)))/(ln(x))=lim_(x->oo)((ln(pi/2-arctan(x)))')/((ln(x))')=lim_(x->oo)(1/(pi/2-arctan(x))*(-1/(1+x^2)))/(1/x)=`

`=lim_(x->oo)(x/(1+x^2))/(arctan(x)-pi/2)`.

On this stage we obtained indeterminate form of type `0/0` so we apply L'Hopital's Rule once more:

`lim_(x->oo)(x/(1+x^2))/(arctan(x)-pi/2)=lim_(x->oo)((x/(1+x^2))')/((arctan(x)-pi/2)')=lim_(x->oo)((1-x^2)/((1+x^2)^2))/(1/(1+x^2))=lim_(x->oo)(1-x^2)/(1+x^2)=lim_(x->oo)(x^2(-1+1/x^2))/(x^2(1-1/x^2))=`

`=lim_(x->oo)(-1+1/x^2)/(1+1/x^2)=-1`.

So, `lim_(x->oo)(pi/2-arctan(x))^(1/(ln(x)))=lim_(x->oo)y=lim_(x->oo)e^(ln(y))=e^(lim_(x->oo)ln(y))=e^(-1)=1/e`.