Welcome back for two more examples
of determining limits using L'Hopital's Rule,
where these two examples involve exponential functions.
L'Hopital's Rule uses derivatives to evaluate limits
involving indeterminant forms,
where the most common indeterminant form
is zero divided by zero,
but these are also considered indeterminant forms.
So if we have a limit that looks like this,
that involves an indeterminant form,
this is equal to the limit as x approaches c,
of f prime of x divided by g prime of x.
So again, if our limit involves an indeterminant form,
we can find the derivative of the numerator,
and the derivative of the denominator,
and see if we can find this limit instead.
So looking at our first example,
notice as x approaches zero,
five to the power of x would approach one,
three to the power of x would approach one,
and one minus one is equal to zero,
and as x approaches zero,
the denominator approaches zero as well,
so this is the indeterminant form of zero divided by zero,
and therefore we can apply L'Hopital's Rule
to help us determine this limit.
But before we do this,
let's look at the graph of this quotient function.
Here's the graph.
Notice how it looks like exponential growth,
but the function is undefined at x equals zero,
but as we approach x from the right, and the left,
we can tell graphically the limit is going to exist.
So let's go ahead and find it using L'Hopital's Rule.
The derivative formulas that we need are given here below,
notice how these first two do include the Chain Rule.
So because this limit involves an indeterminant form,
this limit is equal to the limit as x approaches zero,
of the derivative of five to the power of x,
which should be natural log of the base,
times five to the power of x,
minus the derivative of three to the power of x,
which should be natural log three,
times three to the power of x,
divided by the derivative of x which is equal to one.
And now, because our new function
is continuous around x equals zero,
we can find this limit by performing direct substitution.
Notice how I can just drop the denominator.
We would have natural log five,
times five to the zero,
minus natural log three, times three to the zero,
anything to the zero power is equal to one,
so this leaves us with natural log five,
minus natural log three,
and using our property of logarithms,
we can combine these two logs,
by taking the quotient of the number parts,
meaning this is equal to natural log 5/3.
Before I go back and take a look at the graph,
if we convert this to a decimal,
it would be approximately 0.5108.
So notice how this value is very close to 1/2.
And going back to our graph,
notice how it does appear, that as x approaches zero,
the function value is approaching something close to 1/2,
which is actually natural log of 5/3.
Now let's take a look at our second example,
notice as x approaches zero, denominator approaches zero,
and if we take a look at the numerator,
e to the x would approach e to the zero
which is equal to one,
x approaches zero, so this is approaching one,
this is approaching zero, this is minus one,
and therefore the numerator is also approaching zero,
so we have an indeterminant form,
and therefore we can apply L'Hopital's Rule.
So this limit is equal to the limit as x approaches zero,
of the derivative of the numerator,
divided by the derivative of the denominator.
The derivative of e to the x is e to the x,
the derivative of x is one,
the derivative of negative one is zero,
so e to the x plus one, is the derivative of the numerator,
and the derivative of four x is equal to four.
This new quotient is continuous around x equals zero,
so we can find this limit by performing direct substitution,
so we'd have e to the zero plus one divided by four,
e to the zero is equal to one, one plus one is two,
so this is equal to 2/4, which simplifies to 1/2.
I hope these two examples were helpful.