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.