Other Indeterminate Forms

Example 1. Find $\lim_{{{x}\to{{0}}^{+}}}{x}{\ln{{\left({x}\right)}}}$.

Since ${x}\to{0}$ and ${\ln{{\left({x}\right)}}}\to-\infty$ when ${x}\to{{0}}^{+}$ this is indeterminate form of type ${0}\cdot\infty$.

Representing product as quotient we obtain indeterminate form of type $\frac{{\infty}}{{\infty}}$ and therefore can apply Second L'Hospital's Rule:

$\lim_{{{x}\to{{0}}^{+}}}{x}{\ln{{\left({x}\right)}}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{{\ln{{\left({x}\right)}}}}}{{\frac{{1}}{{x}}}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{{\left({\ln{{\left({x}\right)}}}\right)}'}}{{{\left(\frac{{1}}{{x}}\right)}'}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{\frac{{1}}{{x}}}}{{-\frac{{1}}{{{x}}^{{2}}}}}=$

$=\lim_{{{x}\to{{0}}^{+}}}-{x}={0}$.

Note, that we could represent product as indeterminate form of type $\frac{{0}}{{0}}$: $\lim_{{{x}\to{{0}}^{+}}}{x}{\ln{{\left({x}\right)}}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{x}}{{\frac{{1}}{{{\ln{{\left({x}\right)}}}}}}}$. 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.

Example 2. Compute $\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left({\sec{{\left({x}\right)}}}-{\tan{{\left({x}\right)}}}\right)}$.

Note that ${\sec{{\left({x}\right)}}}\to\infty$ and ${\tan{{\left({x}\right)}}}\to\infty$ as ${x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-{}}}$. Therefore we have indeterminate for of type $\infty-\infty$.

Here we can use a common denominator:

$\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left({\sec{{\left({x}\right)}}}-{\tan{{\left({x}\right)}}}\right)}=\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left(\frac{{1}}{{{\cos{{\left({x}\right)}}}}}-\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}\right)}=\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}\frac{{{1}-{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}$.

We can use L'Hopital's rule now, because $\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left({1}-{\sin{{\left({x}\right)}}}\right)}={0}$ and $\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\cos{{\left({x}\right)}}}={0}$:

$\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}\frac{{{1}-{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}=\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}\frac{{{\left({1}-{\sin{{\left({x}\right)}}}\right)}'}}{{{\left({\cos{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}\frac{{-{\cos{{\left({x}\right)}}}}}{{-{\sin{{\left({x}\right)}}}}}={0}$.

Example 3. Calculate $\lim_{{{x}\to{0}}}{\left({{\cot}}^{{2}}{\left({x}\right)}-\frac{{1}}{{{x}}^{{2}}}\right)}$.

Clearly we have indeterminate form of type $\infty-\infty$ here.

Let's simplify this expression: ${{\cot}}^{{2}}{\left({x}\right)}-\frac{{1}}{{{x}}^{{2}}}=\frac{{{{\cos}}^{{2}}{\left({x}\right)}}}{{{{\sin}}^{{2}}{\left({x}\right)}}}-\frac{{1}}{{{x}}^{{2}}}=\frac{{{{x}}^{{2}}{{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({x}\right)}}}{{{{x}}^{{2}}{{\sin}}^{{2}}{\left({x}\right)}}}=\frac{{{\left({x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}\right)}{\left({x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}\right)}}}{{{{x}}^{{2}}{{\sin}}^{{2}}{\left({x}\right)}}}=$

$=\frac{{{x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}}}{{{x}}}\cdot\frac{{{x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{x}{{\sin}}^{{2}}{\left({x}\right)}}}$.

We can found limit of first factor very easy: $\lim_{{{x}\to{0}}}\frac{{{x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}}}{{x}}=\lim_{{{x}\to{0}}}{\left({\cos{{\left({x}\right)}}}+\frac{{{\sin{{\left({x}\right)}}}}}{{x}}\right)}={\cos{{\left({0}\right)}}}+{1}={2}$.

Limit of second factor is indeterminate form of type $\frac{{0}}{{0}}$ so we can apply L'Hopital's Rule:

$\lim_{{{x}\to{0}}}\frac{{{x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{x}{{\sin}}^{{2}}{\left({x}\right)}}}=\lim_{{{x}\to{0}}}\frac{{{\left({x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}\right)}'}}{{{\left({x}{{\sin}}^{{2}}{\left({x}\right)}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{{\cos{{\left({x}\right)}}}-{x}{\sin{{\left({x}\right)}}}-{\cos{{\left({x}\right)}}}}}{{{{\sin}}^{{2}}{\left({x}\right)}+{2}{x}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}}}=$

$=\lim_{{{x}\to{0}}}\frac{{-{x}}}{{{\sin{{\left({x}\right)}}}+{2}{x}{\cos{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{-{1}}}{{\frac{{{\sin{{\left({x}\right)}}}}}{{x}}+{2}{\cos{{\left({x}\right)}}}}}=\frac{{-{1}}}{{{1}+{2}\cdot{\cos{{\left({0}\right)}}}}}=-\frac{{1}}{{3}}$.

Therefore,

$\lim_{{{x}\to{0}}}{\left({{\cot}}^{{2}}{\left({x}\right)}-\frac{{1}}{{{x}}^{{2}}}\right)}=\lim_{{{x}\to{0}}}{\left(\frac{{{x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}}}{{{x}}}\cdot\frac{{{x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{x}{{\sin}}^{{2}}{\left({x}\right)}}}\right)}=$

$=\lim_{{{x}\to{0}}}\frac{{{x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{x}}}\cdot\lim_{{{x}\to{0}}}\frac{{{x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}}}{{{x}{{\sin}}^{{2}}{\left({x}\right)}}}={2}\cdot{\left(-\frac{{1}}{{3}}\right)}=-\frac{{2}}{{3}}$.

Example 4. Find $\lim_{{{x}\to{{0}}^{+}}}{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}^{{{\cot{{\left({x}\right)}}}}}$.

First, make sure that this is indeterminate limit: since ${1}+{\sin{{\left({2}{x}\right)}}}\to{1}$ and ${\cot{{\left({x}\right)}}}\to\infty$ as ${x}\to{{0}}^{+}$ then this is indeterminate form of type ${{1}}^{{\infty}}$.

Let ${y}={{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}^{{{\cot{{\left({x}\right)}}}}}$ then ${\ln{{\left({y}\right)}}}={\cot{{\left({x}\right)}}}{\ln{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}}=\frac{{{\ln{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}}}}{{{\tan{{\left({x}\right)}}}}}$.

Since ${\ln{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}}\to{0}$ and ${\tan{{\left({x}\right)}}}\to{0}$ as ${x}\to{{0}}^{+}$, we have indeterminate form of type $\frac{{0}}{{0}}$ and can apply L'Hopital's Rule:

$\lim_{{{x}\to{{0}}^{+}}}{\ln{{\left({y}\right)}}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{{\ln{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}}}}{{{\tan{{\left({x}\right)}}}}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{{\left({\ln{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}}\right)}'}}{{{\left({\tan{{\left({x}\right)}}}\right)}'}}=$

$=\lim_{{{x}\to{{0}}^{+}}}\frac{{\frac{{{2}{\cos{{\left({2}{x}\right)}}}}}{{{1}+{\sin{{\left({2}{x}\right)}}}}}}}{{{{\sec}}^{{2}}{\left({x}\right)}}}={2}$.

We found limit of ${\ln{{\left({y}\right)}}}$, but we need limit of ${y}$, so

$\lim_{{{x}\to{{0}}^{+}}}{{\left({1}+{\sin{{\left({2}{x}\right)}}}\right)}}^{{{\cot{{\left({x}\right)}}}}}=\lim_{{{x}\to{{0}}^{+}}}{y}=\lim_{{{x}\to{{0}}^{+}}}{{e}}^{{{\ln{{\left({y}\right)}}}}}={{e}}^{{\lim_{{{x}\to{{0}}^{+}}}{\ln{{\left({y}\right)}}}}}={{e}}^{{2}}$.

Example 5. Calculate $\lim_{{{x}\to{{0}}^{+}}}{{x}}^{{x}}$.

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:

$\lim_{{{x}\to{{0}}^{+}}}{{x}}^{{x}}=\lim_{{{x}\to{{0}}^{+}}}{{e}}^{{{\ln{{\left({{x}}^{{x}}\right)}}}}}=\lim_{{{x}\to{{0}}^{+}}}{{e}}^{{{x}{\ln{{\left({x}\right)}}}}}={{e}}^{{\lim_{{{x}\to{{0}}^{+}}}{x}{\ln{{\left({x}\right)}}}}}={{e}}^{{0}}={1}$ because $\lim_{{{x}\to{{0}}^{+}}}{x}{\ln{{\left({x}\right)}}}={0}$.

Example 6. Find $\lim_{{{x}\to{0}}}{{\left(\frac{{\sin{{\left({x}\right)}}}}{{x}}\right)}}^{{\frac{{1}}{{{1}-{\cos{{\left({x}\right)}}}}}}}$.

Here we have indeterminate form of type ${{1}}^{\infty}$.

Let ${y}={{\left(\frac{{\sin{{\left({x}\right)}}}}{{x}}\right)}}^{{\frac{{1}}{{{1}-{\cos{{\left({x}\right)}}}}}}}$. We can consider the case ${x}>{0}$ (because ${y}$ is odd function).

So, we have that ${\ln{{\left({y}\right)}}}={\ln{{\left({{\left(\frac{{\sin{{\left({x}\right)}}}}{{x}}\right)}}^{{\frac{{1}}{{{1}-{\cos{{\left({x}\right)}}}}}}}\right)}}}=\frac{{1}}{{{1}-{\cos{{\left({x}\right)}}}}}{\ln{{\left(\frac{{\sin{{\left({x}\right)}}}}{{x}}\right)}}}=\frac{{{\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}-{\ln{{\left({x}\right)}}}}}{{{1}-{\cos{{\left({x}\right)}}}}}$.

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

$\lim_{{{x}\to{0}}}{\ln{{\left({y}\right)}}}=\lim_{{{x}\to{0}}}\frac{{{\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}-{\ln{{\left({x}\right)}}}}}{{{1}-{\cos{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{{\left({\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}-{\ln{{\left({x}\right)}}}\right)}'}}{{{\left({1}-{\cos{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{\frac{{{\cos{{\left({x}\right)}}}}}{{{\sin{{\left({x}\right)}}}}}-\frac{{1}}{{x}}}}{{{\sin{{\left({x}\right)}}}}}=$

$=\lim_{{{x}\to{0}}}\frac{{{x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{x}{{\sin}}^{{2}}{\left({x}\right)}}}=-\frac{{1}}{{3}}$ (as was found in Example 3).

So, $\lim_{{{x}\to{0}}}{y}={{e}}^{{-\frac{{1}}{{3}}}}=\frac{{1}}{{{\sqrt[{{3}}]{{{e}}}}}}$.

Example 7. Find $\lim_{{{x}\to\infty}}{{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}^{{\frac{{1}}{{\ln{{\left({x}\right)}}}}}}$.

Since $\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\to{0}$ and $\frac{{1}}{{{\ln{{\left({x}\right)}}}}}\to{0}$ as ${x}\to\infty$, we have indeterminate form of type ${{0}}^{{0}}$.

Let ${y}={{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}^{{\frac{{1}}{{{\ln{{\left({x}\right)}}}}}}}$, then ${\ln{{\left({y}\right)}}}={\ln{{\left({{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}^{{\frac{{1}}{{{\ln{{\left({x}\right)}}}}}}}\right)}}}=\frac{{1}}{{{\ln{{\left({x}\right)}}}}}{\ln{{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}}$.

So, $\lim_{{{x}\to\infty}}{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}\frac{{{\ln{{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}}}}{{{\ln{{\left({x}\right)}}}}}=\lim_{{{x}\to\infty}}\frac{{{\left({\ln{{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}}\right)}'}}{{{\left({\ln{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{\frac{{1}}{{\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}}}\cdot{\left(-\frac{{1}}{{{1}+{{x}}^{{2}}}}\right)}}}{{\frac{{1}}{{x}}}}=$

$=\lim_{{{x}\to\infty}}\frac{{\frac{{x}}{{{1}+{{x}}^{{2}}}}}}{{{\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}}}}$.

On this stage we obtained indeterminate form of type $\frac{{0}}{{0}}$ so we apply L'Hopital's Rule once more:

$\lim_{{{x}\to\infty}}\frac{{\frac{{x}}{{{1}+{{x}}^{{2}}}}}}{{{\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}}}}=\lim_{{{x}\to\infty}}\frac{{{\left(\frac{{x}}{{{1}+{{x}}^{{2}}}}\right)}'}}{{{\left({\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{\frac{{{1}-{{x}}^{{2}}}}{{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}}}}{{\frac{{1}}{{{1}+{{x}}^{{2}}}}}}=\lim_{{{x}\to\infty}}\frac{{{1}-{{x}}^{{2}}}}{{{1}+{{x}}^{{2}}}}=\lim_{{{x}\to\infty}}\frac{{{{x}}^{{2}}{\left(-{1}+\frac{{1}}{{{x}}^{{2}}}\right)}}}{{{{x}}^{{2}}{\left({1}-\frac{{1}}{{{x}}^{{2}}}\right)}}}=$

$=\lim_{{{x}\to\infty}}\frac{{-{1}+\frac{{1}}{{{x}}^{{2}}}}}{{{1}+\frac{{1}}{{{x}}^{{2}}}}}=-{1}$.

So, $\lim_{{{x}\to\infty}}{{\left(\frac{\pi}{{2}}-{\operatorname{arctan}{{\left({x}\right)}}}\right)}}^{{\frac{{1}}{{{\ln{{\left({x}\right)}}}}}}}=\lim_{{{x}\to\infty}}{y}=\lim_{{{x}\to\infty}}{{e}}^{{{\ln{{\left({y}\right)}}}}}={{e}}^{{\lim_{{{x}\to\infty}}{\ln{{\left({y}\right)}}}}}={{e}}^{{-{1}}}=\frac{{1}}{{e}}$.