Understanding Limits and L’Hospital’s Rule

It’s Professor Dave, let’s discuss L’Hospital’s
Rule. Over the past few tutorials we’ve learned
all about differentiation and its applications. With differential calculus more or less wrapped
up, it is just about time to move on to integral calculus, but before that, let’s revisit
limits for a moment. When examining these, we saw that we ran into
problems when trying to evaluate limits for functions where both the numerator and denominator
approached zero. This limit is indeterminate, meaning we just
don’t know what it’s doing. Zero over anything is zero, but anything over
zero is undefined, so we just can’t interpret this result. Luckily there is a rule we can apply in such a case. This is called L’Hospital’s rule. This rule works as follows. Say we have some expression in the form of
f of x over g of x, and we are interested in the limit as x approaches some value, A.
If both of these functions are differentiable, and both of them approach zero as x approaches
A, this indeterminate form can be evaluated if we take the derivative of each function. In other words, the limit of f of x over g
of x as x approaches A is equal to the limit of f prime of x over g prime of x as x approaches A. Let’s apply this to one of the functions
we looked at when we first learned about limits. If we look at sine of x over x, and we want
to know what happens as x approaches zero, it seems impossible, as both sine of x and
x approach zero as x approaches zero. But using this rule, let’s just take the
derivative of the top and bottom. Sine x becomes cosine x, and x becomes one. Now let’s just find the limit of cosine
x as x approaches zero. The cosine of zero is one, so this limit equals one. We got this answer previously, by plugging
values into the function and seeing what happened as we got closer and closer to one, but I
think everyone would agree that using L’Hospital’s rule got us to the answer much faster. Let’s try a couple more for practice. How about the limit of e to the x over x cubed,
as x approaches infinity? Now this one doesn’t give us zero over zero,
it gives us infinity over infinity, which is still an indeterminate form, so let’s
use L’Hospital’s rule. Taking the derivative of the top, this will
remain e to the x. Taking the derivative of the bottom we get
three x squared. This is still infinity over infinity unfortunately. But here’s a great thing about this rule,
why not just take the derivative again? If the limit of the quotient of the functions
is the same as the limit of the quotient of the derivatives, then this truth will hold
no matter how many times we take the derivative. We go again, and e to the x stays as it is
again, while on the bottom we get six x. This is still infinity over infinity, so let’s
go one more time. We get e to the x over six, and finally we
can see that this will simply be infinity. Remember, this rule only applies to limits
of indeterminate form, so zero over zero, or some kind of infinity over some kind of infinity. If either the numerator or denominator is
not one of these, we can’t use the rule, so always check for that first. For example, take something like x plus sine
x over x plus cosine x. We could be tempted to use the rule, but we
must note that while the numerator goes to zero, since sine of zero is zero, the denominator
actually goes to one, because cosine of zero is one. That means zero over one, which is zero, and
we did not have to use L’Hospital’s rule to get there. In fact, if we were to use the rule here,
we would actually get an incorrect answer, as we must have an indeterminate form for
the rule to work. Lastly let’s look at some slightly different
forms of the rule. Say we have an indeterminate product, or the
limit of a product of functions where one goes to zero and the other goes to positive
or negative infinity. Again, we can’t directly determine what
is going on here. In such a case, we can rewrite FG as either
F over (one over G) or G over (one over F). Now we have a form for which we can apply
the rule that we already know. For example, take x times the natural log
of x, as x approaches zero from the positive direction. Naturally, x approaches zero, but the natural
log of x will approach negative infinity. Well let’s write this as one function over
the reciprocal of the other. How about natural log of x over (one over x). Now we can take the derivatives. Natural log of x becomes one over x, and x
to the negative one becomes negative x to the negative two. Let’s multiply the numerator by the reciprocal
of the denominator, and we simply get negative x, so the limit must equal zero. We can do something similar with indeterminate
differences. If the limit of a difference of functions
gives infinity minus infinity, it is impossible to know which infinity wins, to give infinity
or zero as a result. But depending on the functions, we can do
algebra to convert this into a quotient, whether by trigonometric identities or by getting
a common denominator, so that we can apply L’Hospital. Lastly, we can also have indeterminate powers. That’s when a function is raised to the
power of another function, and we end up with something difficult to interpret. This includes zero to the zero, infinity to
the zero, or one to the infinity. In such cases, we can set the expression equal
to y and take the natural log of both sides, so that natural log of y equals natural log
of this expression. This allows us to bring the exponent down
here to the front. Then, so that we have just y instead of LN
y, let’s realize that by the definition of logs, y must equal e to the power of all this. When we take the limit of this, we may now
be able to evaluate it. For example, how about x to the x power as
x approaches zero from the positive direction. This would give us zero to the zero, which
is no good. But if we do the steps outlined a moment ago,
we end up with e raised to the power of the quantity (x times natural log of x). In the previous example, we showed that x
times the natural log of x goes to zero as x approaches zero from the positive direction,
so this limit becomes e to the zero, which is one. So with that, we now understand L’Hospital’s
rule, and we have come full circle with limits and differentiation. We are now truly ready to move on to the other
side of the calculus coin, which is integration, but first, let’s check comprehension.

26 thoughts on “Understanding Limits and L’Hospital’s Rule

  1. A must watch Calculus series for all who are studying this subject. Clear presentation that includes organized content in a textbook format with intelligent, concise and step by step explanation of concepts and worked out examples. I enjoy learning from these videos. Thanks.

  2. Thanks for the online lecture man!
    My teacher at school is so hopeless, she couldn't taught L'Hospital because there's no L'Hospital explanation on textbooks

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