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Indeterminate Form of the Type 00\frac{0}{0}

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We have already talked about the indeterminate forms of the type 00\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 00\frac{{0}}{{0}}.

For example, consider the function f(x)=ex−1x{f{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}-{1}}}{{x}}. We want to analyze the behavior of this function near x=0{x}={0}. In other words, we want to calculate lim⁡x→0ex−1x\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{1}}}{{x}}. Since lim⁡x→0ex−1=0\lim_{{{x}\to{0}}}{{e}}^{{x}}-{1}={0} and lim⁡x→0x=0\lim_{{{x}\to{0}}}{x}={0}, we have an indeterminate form of the type 00\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(x){f{{\left({x}\right)}}} and g(x){g{{\left({x}\right)}}} are differentiable on (a,b]{\left({a},{b}\right]} and g′(x)≠0{g{'}}{\left({x}\right)}\ne{0} on (a,b].{\left({a},{b}\right]}. If lim⁡x→af(x)=0\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={0} and lim⁡x→ag(x)=0\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}={0}, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\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→a{x}\to{a}" can be replaced by any of the following symbols: x→a+{x}\to{{a}}^{+}, x→a−{x}\to{{a}}^{{-{}}}, x→∞{x}\to\infty, or x→−∞{x}\to-\infty.

Example 1. Find lim⁡x→0ex−1x\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{1}}}{{x}}.

Since lim⁡x→0(ex−1)=e0−1=0\lim_{{{x}\to{0}}}{\left({{e}}^{{x}}-{1}\right)}={{e}}^{{0}}-{1}={0} and lim⁡x→0x=0\lim_{{{x}\to{0}}}{x}={0}, we can apply L'Hopital's rule:

lim⁡x→0ex−1x=lim⁡x→0(ex−1)′x′=lim⁡x→0ex1=e0=1\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→0ex−e−xln⁡(e−x)+x−1\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{{\ln{{\left({e}-{x}\right)}}}+{x}-{1}}}.

Since both the numerator and the denominator approach 00 as x→0{x}\to{0}, we can apply L'Hopital's rule:

lim⁡x→0ex−e−xln⁡(e−x)+x−1=lim⁡x→0(ex+e−x)′(ln⁡(e−x)+x−1)′=lim⁡x→0ex+e−x−1e−x+1=2−1e+1=2ee−1\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→0ex−e−x−2xx−sin⁡(x)\lim_{{{x}\to{0}}}\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}-{2}{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}.

Since both the numerator and the denominator approach 00 as x→0{x}\to{0}, we can apply L'Hopital's rule:

lim⁡x→0ex−e−x−2xx−sin⁡(x)=lim⁡x→0(ex−e−x−2x)′(x−sin⁡(x))′=lim⁡x→0ex+e−x−21−cos⁡(x)\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 00\frac{{0}}{{0}} and apply L'Hopital's rule once more:

lim⁡x→0ex+e−x−21−cos⁡(x)=lim⁡x→0(ex+e−x−2)′(1−cos⁡(x))′=lim⁡x→0ex−e−xsin⁡(x)\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 00\frac{{0}}{{0}}, so we apply L'Hopital's rule once more:

lim⁡x→0ex−e−xsin⁡(x)=lim⁡x→0(ex−e−x)′(sin⁡(x))′=lim⁡x→0ex+e−xcos⁡(x)=e0+e−0cos⁡(0)=2\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→0ex−e−x−2xx−sin⁡(x)=2\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→0sin⁡(x)−xx3\lim_{{{x}\to{0}}}\frac{{{\sin{{\left({x}\right)}}}-{x}}}{{{{x}}^{{3}}}}.

Since sin⁡(x)−x→0{\sin{{\left({x}\right)}}}-{x}\to{0} and x3→0{{x}}^{{3}}\to{0} as x→0{x}\to{0}, we have that

lim⁡x→0sin⁡(x)−xx3=lim⁡x→0(sin⁡(x)−x)′(x3)′=lim⁡x→0cos⁡(x)−13x2\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⁡(x)−1→0{\cos{{\left({x}\right)}}}-{1}\to{0} and 3x2→0{3}{{x}}^{{2}}\to{0} as x→0{x}\to{0}, we again have an indeterminate form of the type 00\frac{{0}}{{0}} and apply L'Hopital's rule once more:

lim⁡x→0(cos⁡(x)−1)′(3x2)′=lim⁡x→0−sin⁡(x)6x\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 00\frac{{0}}{{0}}, so we apply L'Hopital's rule for the third time:

lim⁡x→0−sin⁡(x)6x=lim⁡x→0(−sin⁡(x))′(6x)′=lim⁡x→0−cos⁡(x)6=−cos⁡(0)6=−16\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→0+(sin⁡(x))−1cos⁡(x)\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→0+(sin⁡(x)−1)′(cos⁡(x))′=lim⁡x→0+cos⁡(x)−sin⁡(x)=−∞\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→0+cos⁡(x)=1\lim_{{{x}\to{{0}}^{+}}}{\cos{{\left({x}\right)}}}={1} and we don't have an indeterminate form.

In fact, lim⁡x→0sin⁡(x)−1cos⁡(x)=sin⁡(0)−1cos⁡(0)=−1\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→2x2−4x−2\lim_{{{x}\to{2}}}\frac{{{{x}}^{{2}}-{4}}}{{{x}-{2}}}.

Applying L'Hopital's rule gives lim⁡x→2x2−4x−2=lim⁡x→2(x2−4)′(x−2)′=lim⁡x→22x1=4\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→2(x−2)(x+2)x−2=lim⁡x→2(x+2)=4\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→0tan⁡(x)−xx−sin⁡(x)\lim_{{{x}\to{0}}}\frac{{{\tan{{\left({x}\right)}}}-{x}}}{{{x}-{\sin{{\left({x}\right)}}}}}.

We have an indeterminate form of the type 00\frac{{0}}{{0}} and so can apply L'Hopital's rule:

lim⁡x→0tan⁡(x)−xx−sin⁡(x)=lim⁡x→0(tan⁡(x)−x)′(x−sin⁡(x))′=lim⁡x→01cos⁡2(x)−11−cos⁡(x)\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 00\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: 1cos⁡2(x)−1cos⁡(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)\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→01cos⁡2(x)−11−cos⁡(x)=lim⁡x→01+cos⁡(x)cos⁡2(x)=1+cos⁡(0)cos⁡(0)=2\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→0xe2x+xex−2e2x+2ex(ex−1)3\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→0xe2x+xex−2e2x+2ex(ex−1)3=lim⁡x→0e2x+2xe2x+ex+xex−4e2x+2ex3ex(ex−1)2=\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→02xex−3ex+3+x3(ex−1)2==\lim_{{{x}\to{0}}}\frac{{{2}{x}{{e}}^{{x}}-{3}{{e}}^{{x}}+{3}+{x}}}{{{3}{{\left({{e}}^{{x}}-{1}\right)}}^{{2}}}}=

=lim⁡x→02ex+2xex−3ex+16ex(ex−1)=lim⁡x→02x−1+e−x6(ex−1)=lim⁡x→02−e−x6ex=16=\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}}.