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l’Hopital’s Rule

April 10th, 2009 |

Author: Merrin

l’Hopital’s Rule is extremely useful in calculating limits which are stuck in an indeterminate form. As a reminder, an indeterminate form is a quantity calculated in the collision of two limits that can take on any value depending on the strengths of the individual limits. Some common indeterminate forms encountered in calculus are 0/0, $latex \infty/\infty $, $latex \infty -\infty$, $latex 0\cdot \infty$, $latex \infty ^{0}$, $latex \infty ^{0}$ , $latex 1^{\infty }$. Many of these indeterminate forms can be dealt with in one way or another with l’Hopital’s rule.

Theorem 1. l’Hopital’s Rule states that if a limit has one of the following two indeterminate forms,

$\mathop{\lim }\limits_{x\to a} \cfrac{f(x)}{g(x)} =\cfrac{\pm \infty }{\pm \infty } \qquad \mathop{\lim }\limits_{x\to a} \cfrac{f(x)}{g(x)} =\cfrac{0}{0}$

Then attempt to evaluate the limit by taking the derivative of the numerator and denominator according to this formula.

$\mathop{\lim }\limits_{x\to a} \cfrac{f(x)}{g(x)} =\mathop{\lim }\limits_{x\to a} \cfrac{f'(x)}{g'(x)}$

If the result of differentiation gives another one of these indeterminate forms then the rule can be applied again. In your work, indicate the indeterminate form is of the right type before differentiating the numerator and the denominator. Don’t go crazy and start differentiating any old expression. It has to be one of the kinds specified in the theorem, zero over zero or infinity over infinity.

Example 1. Find the following trigonometric limit.

$\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x}$

Solution 1. This limit is one of our old friends that took us some trouble to find with the squeeze theorem. l’Hopital’s rule makes short work of it. This limit is an indeterminate form that l’Hopital’s Rule can be applied to, zero over zero.

$\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x} =\cfrac{0}{0} =\mathop{\lim }\limits_{x\to 0} \cfrac{(\sin x)'}{x'} =\mathop{\lim }\limits_{x\to 0} \cfrac{\cos x}{1} =1$

Example 2. Find the following limit.

$\mathop{\lim }\limits_{x\to 0} \cfrac{1-\cos x}{x^{2} }$

Solution 2. This limit is an indeterminate form that l’Hopital’s Rule can be applied to. Solving this limit uses l’Hopital’s Rule twice including the result from above.

$\mathop{\lim }\limits_{x\to 0} \cfrac{1-\cos x}{x^{2} } =\cfrac{0}{0} =\mathop{\lim }\limits_{x\to 0} \cfrac{(1-\cos x)'}{(x^{2} )'} =\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{2x} =\cfrac{1}{2}$

Example 3. Find the following limit by applying l’Hopital’s Rule.

$\mathop{\lim }\limits_{x\to \infty } \cfrac{x^{2} }{e^{x} }$

Solution 3. This limit is of the second indeterminate form allowed for using l’Hopital’s Rule. When the indeterminate form occurs we differentiate the numerator and the denominator.

$\mathop{\lim }\limits_{x\to \infty } \cfrac{x^{2} }{e^{x} } =\cfrac{\infty }{\infty } =\mathop{\lim }\limits_{x\to \infty } \cfrac{2x}{e^{x} } =\cfrac{\infty }{\infty } =\mathop{\lim }\limits_{x\to \infty } \cfrac{2}{e^{x} } =0$

In general, for any positive integer to the nth power the limit can be found by using l’Hopital’s Rule n times. Remember that $latex e^{x} $ is the boss over $latex x^{n} $.

$\mathop{\lim }\limits_{x\to \infty } \cfrac{x^{n} }{e^{x} } =0$

Example 4. Find the following limit L.

Solution 4.

$L=\mathop{\lim }\limits_{x\to \infty } \cfrac{\ln x}{x} =\cfrac{\infty }{\infty } =\mathop{\lim }\limits_{x\to \infty }\cfrac{1/x}{1} =0$

Here we see that x is the boss over ln x.

The other indeterminate forms are incompatible with l’Hopital’s Rule at first sight, but some manipulations allow us to use l’Hopital’s Rule with them. The first indeterminate form we consider is

$\mathop{\lim }\limits_{x\to a} [f(x)-g(x)]=\infty -\infty$

If f(x) and g(x) are made to have a common denominator, perhaps the numerator and denominator will fit the form of l’Hopital’ s rule. Multiplication by a conjugate function may also make a useful fraction.

Example 5. Find the limit of

$L=\mathop{\lim }\limits_{x\to \infty } \left(\sqrt{x} -\sqrt{x-1} \right)=\infty -\infty$

Solution 5. Combine the two pieces into one term with a conjugate term.

$L = {\mathop{\lim }\limits_{x\to \infty }} \left( \cfrac{\sqrt{x} -\sqrt{x-1} }{\sqrt{x} +\sqrt{x-1} } \right) \left( \sqrt{x} +\sqrt{x-1} \right)$

Combine the two numerators and there is some cancellation.

$L = \mathop{\lim }\limits_{x\to \infty } \cfrac{1}{2\sqrt{x}} =0$

Example 6. Find the limit of

$L = \mathop{\lim }\limits_{x \to 0} \left(\cfrac{1}{\tan x} -\cfrac{1}{\sin x} \right)$

Solution 6. The current form of the limit is an indeterminate form of infinity minus infinity. We want to combine the two terms with the same denominator so we can use l’Hopital’s Rule hopefully.

$L ={\mathop{\lim }\limits_{x\to 0} \cfrac{(\sin x)-(\tan x)}{(\tan x)(\sin x)} =\cfrac{0}{0} } \\ \\ ={\mathop{\lim }\limits_{x\to 0} \cfrac{\cos x-(\sec x)^{2} }{(\sec x)^{2} \sin x+\tan x\cos x} } \\ \\= {\cfrac{(1-1)}{0+0} =\cfrac{0}{0} } \\ \\= {\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x-2(\sec x)^{2} \tan x}{(\sec x)^{2} (\cos x+2\tan x\sin x+\cos x)+\tan x\sin x} } \\ \\= {\cfrac{0}{2} =0}$

There are several other indeterminate forms we wish to consider where l’Hopital’s Rule does not apply directly, but with a manipulation it is usable. Consider the indeterminate form.

$L=\mathop{\lim }\limits_{x\to a} f(x)g(x)=0\cdot \infty$

There are two options. Flip the 0 part, f(x), into the denominator or flip the infinity part, g(x) into the denominator.

${L=0\cdot \infty =\cfrac{\infty }{1/0} =\cfrac{\infty }{\infty } } \\ \\ {L=0\cdot \infty =\cfrac{0}{1/\infty } =\cfrac{0}{0} }$

In each case, flipping a factor into the denominator puts the limit into a form that l’Hopital’s Rule is ready to operate on.

Example 7. Consider the commonly occurring limit

$L=\mathop{\lim }\limits_{x\to 0^{+} } x\ln x$

Solution 7. We flip the factor x into the denominator.

$L=\mathop{\lim }\limits_{x\to 0^{+} } \cfrac{\ln x}{1/x} =\cfrac{-\infty }{\infty } =\mathop{\lim }\limits_{x\to 0^{+} } \cfrac{1/x}{-1/x^{2} } =-\mathop{\lim }\limits_{x\to 0^{+} } x=0$

Flipping the factor ln x into the denominator only complicates matters because the derivative of its reciprocal is nasty.

Now we will consider indeterminate forms that are power laws.

Example 8. First we know that any number to the 0 power is zero, well what about $latex 0^0$? Traditionally this number is undefined, but we can pose the question as a limit and see what we should define it as.

$L=\mathop{\lim }\limits_{x\to 0^{+} } x^{x}$

Solution 8. Now how are we to proceed? One method is to take the logarithm of the function, work on that, and then exponentiate back at the end. Since we just did the relevant limit we can quote the result.

$\ln L=\mathop{\lim }\limits_{x\to 0^{+} } x\ln x=0$

Now we exponentiate back and find

$e^{\ln L} =e^{0} =L$

$\mathop{\lim }\limits_{x\to 0^{+} } x^{x} =1$

We make have made a case that $latex 0^{0} =1$ should be the definition.

All limits of the form $latex a^{b}$ where $latex a,b\in \{0,1,\infty ,-\infty \} $ can be handled by taking the logarithm and evaluating the limit if possible. Here is one last example.

Example 9. Find the limit L.

$L=\mathop{\lim }\limits_{x\to \infty } x^{1/x}$

Solution 9. We notice a power law so we proceed by taking the logarithm.

${\ln L} = {\mathop{\lim }\limits_{x\to \infty } \cfrac{\ln x}{x} =\cfrac{\infty }{\infty } =\mathop{\lim }\limits_{x\to \infty } \cfrac{1/x}{1} =0} \\ \\{L}= {e^{0} =1}$

In this case, $latex \infty^{1/\infty}=1$

We are now very well equipped to solve some difficult limits. All it takes is practice to calculate limits and applying straightforward methods. Usually one relies heavily on l’Hopital’s rule but don’t forget about multiplying by conjugate factors and other methods. There is one last method to find limits we have yet to encounter in this book which is to use series expansion of a function. This method is especially useful in situations where l’Hopital’s Rule is used over and over with no end in sight.

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