# 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)) .

1. 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 ).
2. lim_(x->a)f(x)=oo and lim_(x->a)g(x)=0 give indeterminate form of type oo^0.
3. 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:

1. Taking the natural logarithm: let y=(f(x))^(g(x)) then ln(y)=ln((f(x))^(g(x)))=g(x)ln(f(x)).
2. 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.