Indeterminate Form of the Type $$$\frac{0}{0}$$$
We have already talked about the indeterminate forms of the type $$$\frac{{0}}{{0}}$$$. We have studied the rational functions and performed algebraic manipulations to get rid of indetermination.
However, there are functions that are not rational but still have an indeterminate form of the type $$$\frac{{0}}{{0}}$$$.
For example, consider the function $$${f{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}-{1}}}{{x}}$$$. We want to analyze the behavior of this function near $$${x}={0}$$$. In other words, we want to calculate $$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{1}}}{{x}}$$$. Since $$$\lim_{{{x}\to{0}}}{{e}}^{{x}}-{1}={0}$$$ and $$$\lim_{{{x}\to{0}}}{x}={0}$$$, we have an indeterminate form of the type $$$\frac{{0}}{{0}}$$$. But here we can't perform algebraic manipulations to simplify the expression and get rid of indetermination. We just can't simplify it.
Luckily, there is a powerful method that allows us to find such a type of limit.
First L'Hopital’s Rule. Suppose that $$${f{{\left({x}\right)}}}$$$ and $$${g{{\left({x}\right)}}}$$$ are differentiable on $$${\left({a},{b}\right]}$$$ and $$${g{'}}{\left({x}\right)}\ne{0}$$$ on $$${\left({a},{b}\right]}.$$$ If $$$\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={0}$$$ and $$$\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}={0}$$$, then $$$\lim_{{{x}\to{a}}}\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}=\lim_{{{x}\to{a}}}\frac{{{f{'}}{\left({x}\right)}}}{{{g{'}}{\left({x}\right)}}}$$$ if the limit on the right side exists (or is $$$\infty$$$ or $$$-\infty$$$ ).
It is especially important to verify the conditions before using L'Hopital's rule.
The first L'Hopital’s rule is also valid for the one-sided limits and for the limits at infinity or negative infinity; that is, "$$${x}\to{a}$$$" can be replaced by any of the following symbols: $$${x}\to{{a}}^{+}$$$, $$${x}\to{{a}}^{{-{}}}$$$, $$${x}\to\infty$$$, or $$${x}\to-\infty$$$.
Example 1. Find $$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{1}}}{{x}}$$$.
Since $$$\lim_{{{x}\to{0}}}{\left({{e}}^{{x}}-{1}\right)}={{e}}^{{0}}-{1}={0}$$$ and $$$\lim_{{{x}\to{0}}}{x}={0}$$$, we can apply L'Hopital's rule:
$$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{1}}}{{x}}=\lim_{{{x}\to{0}}}\frac{{{\left({{e}}^{{x}}-{1}\right)}'}}{{{x}'}}=\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}}}{{1}}={{e}}^{{0}}={1}$$$.
Let's proceed to another example.
Example 2. Find $$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{\ln{{\left({e}-{x}\right)}}}+{x}-{1}}}$$$.
Since both the numerator and the denominator approach $$$0$$$ as $$${x}\to{0}$$$, we can apply L'Hopital's rule:
$$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{\ln{{\left({e}-{x}\right)}}}+{x}-{1}}}=\lim_{{{x}\to{0}}}\frac{{{\left({{e}}^{{x}}+{{e}}^{{-{x}}}\right)}'}}{{{\left({\ln{{\left({e}-{x}\right)}}}+{x}-{1}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{-\frac{{1}}{{{e}-{x}}}+{1}}}=\frac{{{2}}}{{-\frac{{1}}{{e}}+{1}}}=\frac{{{2}{e}}}{{{e}-{1}}}$$$.
Sometimes we need to apply L'Hopital's rule more than once.
Example 3. Find $$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}-{2}{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}$$$.
Since both the numerator and the denominator approach $$$0$$$ as $$${x}\to{0}$$$, we can apply L'Hopital's rule:
$$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}-{2}{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}-{2}{x}\right)}'}}{{{\left({x}-{\sin{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}-{2}}}{{{1}-{\cos{{\left({x}\right)}}}}}$$$.
We again obtained an indeterminate form of the type $$$\frac{{0}}{{0}}$$$ and apply L'Hopital's rule once more:
$$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}-{2}}}{{{1}-{\cos{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{{\left({{e}}^{{x}}+{{e}}^{{-{x}}}-{2}\right)}'}}{{{\left({1}-{\cos{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{\sin{{\left({x}\right)}}}}}$$$.
We again obtained an indeterminate form of the type $$$\frac{{0}}{{0}}$$$, so we apply L'Hopital's rule once more:
$$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{\sin{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}'}}{{{\left({\sin{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}+{{e}}^{{-{x}}}}}{{{\cos{{\left({x}\right)}}}}}=\frac{{{{e}}^{{0}}+{{e}}^{{-{0}}}}}{{{\cos{{\left({0}\right)}}}}}={2}$$$.
So, $$$\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}-{2}{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}={2}$$$.
Now, let's do some more work.
Example 4. Find $$$\lim_{{{x}\to{0}}}\frac{{{\sin{{\left({x}\right)}}}-{x}}}{{{{x}}^{{3}}}}$$$.
Since $$${\sin{{\left({x}\right)}}}-{x}\to{0}$$$ and $$${{x}}^{{3}}\to{0}$$$ as $$${x}\to{0}$$$, we have that
$$$\lim_{{{x}\to{0}}}\frac{{{\sin{{\left({x}\right)}}}-{x}}}{{{x}}^{{3}}}=\lim_{{{x}\to{0}}}\frac{{{\left({\sin{{\left({x}\right)}}}-{x}\right)}'}}{{{\left({{x}}^{{3}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{{\cos{{\left({x}\right)}}}-{1}}}{{{3}{{x}}^{{2}}}}$$$.
Since $$${\cos{{\left({x}\right)}}}-{1}\to{0}$$$ and $$${3}{{x}}^{{2}}\to{0}$$$ as $$${x}\to{0}$$$, we again have an indeterminate form of the type $$$\frac{{0}}{{0}}$$$ and apply L'Hopital's rule once more:
$$$\lim_{{{x}\to{0}}}\frac{{{\left({\cos{{\left({x}\right)}}}-{1}\right)}'}}{{{\left({3}{{x}}^{{2}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{-{\sin{{\left({x}\right)}}}}}{{{6}{x}}}$$$.
Again, we have an indeterminate form of the type $$$\frac{{0}}{{0}}$$$, so we apply L'Hopital's rule for the third time:
$$$\lim_{{{x}\to{0}}}\frac{{-{\sin{{\left({x}\right)}}}}}{{{6}{x}}}=\lim_{{{x}\to{0}}}\frac{{{\left(-{\sin{{\left({x}\right)}}}\right)}'}}{{{\left({6}{x}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{-{\cos{{\left({x}\right)}}}}}{{6}}=-\frac{{\cos{{\left({0}\right)}}}}{{6}}=-\frac{{1}}{{6}}$$$.
And one more example.
Example 5. Find $$$\lim_{{{x}\to{{0}}^{+}}}\frac{{{\left({\sin{{\left({x}\right)}}}\right)}-{1}}}{{{\cos{{\left({x}\right)}}}}}$$$.
If we blindly attempt to apply L'Hopital's rule, we will get that $$$\lim_{{{x}\to{{0}}^{+}}}\frac{{{\left({\sin{{\left({x}\right)}}}-{1}\right)}'}}{{{\left({\cos{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{{\cos{{\left({x}\right)}}}}}{{-{\sin{{\left({x}\right)}}}}}=-\infty$$$.
THIS IS WRONG! We can't apply L'Hopital's rule because $$$\lim_{{{x}\to{{0}}^{+}}}{\cos{{\left({x}\right)}}}={1}$$$ and we don't have an indeterminate form.
In fact, $$$\lim_{{{x}\to{0}}}\frac{{{\sin{{\left({x}\right)}}}-{1}}}{{\cos{{\left({x}\right)}}}}=\frac{{{\sin{{\left({0}\right)}}}-{1}}}{{{\cos{{\left({0}\right)}}}}}=-{1}$$$.
Example 5 shows what can go wrong if you use L'Hopital's rule without checking the conditions of the theorem.
Other limits can be found using L'Hopital's rule but are found more easily by means of other methods. So, when evaluating any limit, you should consider other methods before using L'Hopital's rule.
Memorize this face, and let's proceed.
Example 6. Find $$$\lim_{{{x}\to{2}}}\frac{{{{x}}^{{2}}-{4}}}{{{x}-{2}}}$$$.
Applying L'Hopital's rule gives $$$\lim_{{{x}\to{2}}}\frac{{{{x}}^{{2}}-{4}}}{{{x}-{2}}}=\lim_{{{x}\to{2}}}\frac{{{\left({{x}}^{{2}}-{4}\right)}'}}{{{\left({x}-{2}\right)}'}}=\lim_{{{x}\to{2}}}\frac{{{2}{x}}}{{1}}={4}$$$.
But it is more natural to use algebraic manipulations: $$$\lim_{{{x}\to{2}}}\frac{{{\left({x}-{2}\right)}{\left({x}+{2}\right)}}}{{{x}-{2}}}=\lim_{{{x}\to{2}}}{\left({x}+{2}\right)}={4}$$$.
Sometimes we need to combine L'Hopital's rule with algebraic manipulations.
Example 7. Find $$$\lim_{{{x}\to{0}}}\frac{{{\tan{{\left({x}\right)}}}-{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}$$$.
We have an indeterminate form of the type $$$\frac{{0}}{{0}}$$$ and so can apply L'Hopital's rule:
$$$\lim_{{{x}\to{0}}}\frac{{{\tan{{\left({x}\right)}}}-{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{{\left({\tan{{\left({x}\right)}}}-{x}\right)}'}}{{{\left({x}-{\sin{{\left({x}\right)}}}\right)}'}}=\lim_{{{x}\to{0}}}\frac{{\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}-{1}}}{{{1}-{\cos{{\left({x}\right)}}}}}$$$.
We again have an indeterminate form of the type $$$\frac{{0}}{{0}}$$$; so, we can apply L'Hopital's rule once more. However, if we blindly use L'Hopital's rule, we of course will obtain the correct answer, but we need to apply L'Hopital's rule three times (you can check it), and taking the derivatives won't be pleasant.
Instead, we perform some algebraic manipulations: $$$\frac{{\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}-{1}}}{{{\cos{{\left({x}\right)}}}-{1}}}=\frac{{{1}-{{\cos}}^{{2}}{\left({x}\right)}}}{{{{\cos}}^{{2}}{\left({x}\right)}{\left({1}-{\cos{{\left({x}\right)}}}\right)}}}=\frac{{{\left({1}-{\cos{{\left({x}\right)}}}\right)}{\left({1}+{\cos{{\left({x}\right)}}}\right)}}}{{{{\cos}}^{{2}}{\left({x}\right)}{\left({1}-{\cos{{\left({x}\right)}}}\right)}}}=\frac{{{1}+{\cos{{\left({x}\right)}}}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}$$$.
So, $$$\lim_{{{x}\to{0}}}\frac{{\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}-{1}}}{{{1}-{\cos{{\left({x}\right)}}}}}=\lim_{{{x}\to{0}}}\frac{{{1}+{\cos{{\left({x}\right)}}}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}=\frac{{{1}+{\cos{{\left({0}\right)}}}}}{{\cos{{\left({0}\right)}}}}={2}$$$.
So, the algebraic manipulations saved us time and effort.
Let's do our final example.
Example 8. Calculate $$$\lim_{{{x}\to{0}}}\frac{{{x}{{e}}^{{{2}{x}}}+{x}{{e}}^{{x}}-{2}{{e}}^{{{2}{x}}}+{2}{{e}}^{{x}}}}{{{{\left({{e}}^{{x}}-{1}\right)}}^{{3}}}}$$$.
Applying L'Hopital's rule a couple of times and simplifying the result between the applications gives:
$$$\lim_{{{x}\to{0}}}\frac{{{x}{{e}}^{{{2}{x}}}+{x}{{e}}^{{x}}-{2}{{e}}^{{{2}{x}}}+{2}{{e}}^{{x}}}}{{{{\left({{e}}^{{x}}-{1}\right)}}^{{3}}}}=\lim_{{{x}\to{0}}}\frac{{{{e}}^{{{2}{x}}}+{2}{x}{{e}}^{{{2}{x}}}+{{e}}^{{x}}+{x}{{e}}^{{x}}-{4}{{e}}^{{{2}{x}}}+{2}{{e}}^{{x}}}}{{{3}{{e}}^{{x}}{{\left({{e}}^{{x}}-{1}\right)}}^{{2}}}}=$$$
$$$=\lim_{{{x}\to{0}}}\frac{{{2}{x}{{e}}^{{x}}-{3}{{e}}^{{x}}+{3}+{x}}}{{{3}{{\left({{e}}^{{x}}-{1}\right)}}^{{2}}}}=$$$
$$$=\lim_{{{x}\to{0}}}\frac{{{2}{{e}}^{{x}}+{2}{x}{{e}}^{{x}}-{3}{{e}}^{{x}}+{1}}}{{{6}{{e}}^{{x}}{\left({{e}}^{{x}}-{1}\right)}}}=\lim_{{{x}\to{0}}}\frac{{{2}{x}-{1}+{{e}}^{{-{x}}}}}{{{6}{\left({{e}}^{{x}}-{1}\right)}}}=\lim_{{{x}\to{0}}}\frac{{{2}-{{e}}^{{-{x}}}}}{{{6}{{e}}^{{x}}}}=\frac{{1}}{{6}}$$$.