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Indeterminate Forms of Type ∞∞\frac{\infty}{\infty}

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Similarly there are limit of functions that represent indeterminate form of type ∞∞\frac{\infty}{\infty}, but can't be calculated using algebraic manipulations.

However, there is corresponding L'Hopital's Rule that allows to handle indeterminate form of type ∞∞\frac{\infty}{\infty}.

Second L'Hopital’s Rule. Suppose 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)=∞\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}=\infty and lim⁡x→ag(x)=∞\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}=\infty, 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.

Second L'Hopital’s Rule is also valid for one-sided limits and for 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, x→−∞{x}\to-\infty.

Example 1. Find lim⁡x→∞ln⁡(x)x\lim_{{{x}\to\infty}}\frac{{{\ln{{\left({x}\right)}}}}}{{\sqrt{{{x}}}}}.

Since lim⁡x→∞(ln⁡(x))=∞\lim_{{{x}\to\infty}}{\left({\ln{{\left({x}\right)}}}\right)}=\infty and lim⁡x→∞x=∞\lim_{{{x}\to\infty}}\sqrt{{{x}}}=\infty then we can apply L'Hopital's Rule:

lim⁡x→∞ln⁡(x)x=lim⁡x→∞(ln⁡(x))′(x)′=lim⁡x→∞1x12x=lim⁡x→∞2x=0\lim_{{{x}\to\infty}}\frac{{{\ln{{\left({x}\right)}}}}}{{\sqrt{{{x}}}}}=\lim_{{{x}\to\infty}}\frac{{{\left({\ln{{\left({x}\right)}}}\right)}'}}{{{\left(\sqrt{{{x}}}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{\frac{{1}}{{x}}}}{{\frac{{1}}{{{2}\sqrt{{{x}}}}}}}=\lim_{{{x}\to\infty}}\frac{{2}}{{\sqrt{{{x}}}}}={0}.

Sometimes we need to apply L'Hopital's rule more than once.

Example 2. Find lim⁡x→∞exx2\lim_{{{x}\to\infty}}\frac{{{{e}}^{{x}}}}{{{{x}}^{{2}}}}.

Since lim⁡x→∞ex=∞\lim_{{{x}\to\infty}}{{e}}^{{x}}=\infty and lim⁡x→∞x2=∞\lim_{{{x}\to\infty}}{{x}}^{{2}}=\infty then we can use L'Hopital's Rule:

lim⁡x→∞exx2=lim⁡x→∞(ex)′(x2)′=lim⁡x→∞ex2x\lim_{{{x}\to\infty}}\frac{{{{e}}^{{x}}}}{{{{x}}^{{2}}}}=\lim_{{{x}\to\infty}}\frac{{{\left({{e}}^{{x}}\right)}'}}{{{\left({{x}}^{{2}}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{{{e}}^{{x}}}}{{{2}{x}}}.

Since ex→∞{{e}}^{{x}}\to\infty and 2x→∞{2}{x}\to\infty as x→∞{x}\to\infty then we still have indeterminate form of type ∞∞\frac{{\infty}}{{\infty}} and we apply L'Hopital's rule once more:

lim⁡x→∞ex2x=lim⁡x→∞(ex)′(2x)′=lim⁡x→∞ex2=∞\lim_{{{x}\to\infty}}\frac{{{{e}}^{{x}}}}{{{2}{x}}}=\lim_{{{x}\to\infty}}\frac{{{\left({{e}}^{{x}}\right)}'}}{{{\left({2}{x}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{{{e}}^{{x}}}}{{2}}=\infty.

Example 3. Find lim⁡x→∞1+1xx+1\lim_{{{x}\to\infty}}\frac{{{1}+\frac{{1}}{{x}}}}{{{x}+{1}}}.

If we blindly attempt to apply L'Hopital's Rule, we will get that lim⁡x→∞(1+1x)′(x+1)′=lim⁡x→∞1−1x21=1\lim_{{{x}\to\infty}}\frac{{{\left({1}+\frac{{1}}{{x}}\right)}'}}{{{\left({x}+{1}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{{1}-\frac{{1}}{{{x}}^{{2}}}}}{{{1}}}={1}.

THIS IS WRONG! We can't apply L'Hopital's rule because lim⁡x→∞(1+1x)=1\lim_{{{x}\to\infty}}{\left({1}+\frac{{1}}{{x}}\right)}={1} and we don't have indeterminate form.

In fact lim⁡x→∞1+1xx+1=lim⁡x→∞x+1xx+1=lim⁡x→∞1x=0\lim_{{{x}\to\infty}}\frac{{{1}+\frac{{1}}{{x}}}}{{{x}+{1}}}=\lim_{{{x}\to\infty}}\frac{{\frac{{{x}+{1}}}{{x}}}}{{{x}+{1}}}=\lim_{{{x}\to\infty}}\frac{{1}}{{x}}={0}.

Example 3 shows what can go wrong if you use L'Hopital's Rule without thinking checking conditions of theorem.

Now let's see what will be if we ignore condition that limit of ratio of derivatives should exist.

Example 4. Calculate lim⁡x→∞x+sin⁡(x)x\lim_{{{x}\to\infty}}\frac{{{x}+{\sin{{\left({x}\right)}}}}}{{x}}.

We have indeterminate form here, so can apply L'Hopital's Rule:

lim⁡x→∞(x+sin⁡(x))′x′=lim⁡x→∞1+cos⁡(x)1\lim_{{{x}\to\infty}}\frac{{{\left({x}+{\sin{{\left({x}\right)}}}\right)}'}}{{{x}'}}=\lim_{{{x}\to\infty}}\frac{{{1}+{\cos{{\left({x}\right)}}}}}{{1}}.

Since cos⁡(x){\cos{{\left({x}\right)}}} oscillates infinitely many times as x→∞{x}\to\infty then lim⁡x→∞cos⁡(x)\lim_{{{x}\to\infty}}{\cos{{\left({x}\right)}}} doesn't exist. Therefore lim⁡x→∞(1+cos⁡(x))\lim_{{{x}\to\infty}}{\left({1}+{\cos{{\left({x}\right)}}}\right)} doesn't exist.

However, initial limit exist: lim⁡x→∞x+sin⁡(x)x=lim⁡x→∞(1+sin⁡(x)x)=1\lim_{{{x}\to\infty}}\frac{{{x}+{\sin{{\left({x}\right)}}}}}{{x}}=\lim_{{{x}\to\infty}}{\left({1}+\frac{{\sin{{\left({x}\right)}}}}{{x}}\right)}={1}.

So, we need to be sure that limit of ratio of derivative exists, otherwise L'Hopital's Rule is inapplicable.

Other limits can be found using L'Hopital's Rule but are more easily found by other methods. So when evaluating any limit, you should consider other methods before using L'Hopital's Rule.

Example 5. Find lim⁡x→∞x2−42x2−2\lim_{{{x}\to\infty}}\frac{{{{x}}^{{2}}-{4}}}{{{2}{{x}}^{{2}}-{2}}}.

Applying L'Hopital's Rule gives lim⁡x→∞x2−42x2−2=lim⁡x→∞(x2−4)′(2x2−2)′=lim⁡x→∞2x4x=lim⁡x→∞12=12\lim_{{{x}\to\infty}}\frac{{{{x}}^{{2}}-{4}}}{{{2}{{x}}^{{2}}-{2}}}=\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{2}}-{4}\right)}'}}{{{\left({2}{{x}}^{{2}}-{2}\right)}'}}=\lim_{{{x}\to\infty}}\frac{{{2}{x}}}{{{4}{x}}}=\lim_{{{x}\to\infty}}\frac{{1}}{{2}}=\frac{{1}}{{2}}.

But it is more natural to use algebraic manipulations: lim⁡x→∞x2(1−4x2)x2(2−2x2)=lim⁡x→∞1−4x22−2x2=12\lim_{{{x}\to\infty}}\frac{{{{x}}^{{2}}{\left({1}-\frac{{4}}{{{x}}^{{2}}}\right)}}}{{{{x}}^{{2}}{\left({2}-\frac{{2}}{{{x}}^{{2}}}\right)}}}=\lim_{{{x}\to\infty}}\frac{{{1}-\frac{{4}}{{{x}}^{{2}}}}}{{{2}-\frac{{2}}{{{x}}^{{2}}}}}=\frac{{1}}{{2}}.