Video Transcript
Determine the limit as 𝑥
approaches zero of five times 𝑒 to the power of 19𝑥 minus five times 𝑒 to the
power of 10𝑥 all divided by 𝑥.
We’re asked to evaluate the limit
as 𝑥 approaches zero of the quotient of two functions. We can see in our numerator we have
the difference between two exponential functions. And in our denominator, we just
have the function 𝑥. So both our numerator and
denominator are continuous. This means we can attempt to
evaluate this limit by using direct substitution. So we substitute 𝑥 is equal to
zero into the function inside of our limit, giving us five 𝑒 to the power of 19
times zero minus five 𝑒 to the power of 10 times zero all divided by zero.
And of course, we can simplify. 19 times zero and 10 times zero are
both equal to zero, and 𝑒 to the zeroth power is just equal to one. So our numerator simplifies to give
us five minus five. But five minus five is equal to
zero. So by using direct substitution, we
got the indeterminate form of zero divided by zero. This means we’re going to need to
find a different way to evaluate this limit. And since this is the limit of the
quotient of two continuous functions and we know how to differentiate both our
numerator and our denominator and direct substitution gave us the indeterminate form
of zero over zero, we can try using L’Hôpital’s rule.
So let’s recall the following
version of L’Hôpital’s rule. This tells us if 𝑓 and 𝑔 are
differentiable around 𝑥 is equal to 𝑎 and 𝑔 prime of 𝑥 is not equal to zero
around 𝑎, although 𝑔 prime of 𝑎 is allowed to be equal to zero, and both the
limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥
are equal to zero, then L’Hôpital’s rule tells us the limit as 𝑥 approaches 𝑎 of
𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥
divided by 𝑔 prime of 𝑥.
And it’s worth pointing out this is
only true if this limit exists or it’s equal to positive or negative ∞. So L’Hôpital’s rule gives us a
method of turning a question about the limits of the quotient of two functions into
a question about the limits of the quotient of their derivatives.
Let’s see if we can use L’Hôpital’s
rule on the limit given to us in the question. First, we can see our limit is as
𝑥 is approaching zero. So we’ll set our value of 𝑎 equal
to zero. And in fact, we’ll update our
version of L’Hôpital’s rule with 𝑎 set to be zero. Next, we’ll set 𝑓 of 𝑥 to be the
function in our numerator and 𝑔 of 𝑥 to be the function in our denominator. So that’s 𝑓 of 𝑥 is equal to five
𝑒 to the power of 19𝑥 minus five 𝑒 to the power of 10𝑥 and 𝑔 of 𝑥 is 𝑥.
We’re now ready to check our
prerequisites for L’Hôpital’s rule. First, we need to check that both
𝑓 and 𝑔 are differentiable for values of 𝑥 around 𝑥 is equal to zero. And we can see this directly from
the definitions of 𝑓 and 𝑔. First, 𝑓 is the difference between
two exponential functions. Exponential functions are
differentiable for all real values of 𝑥. So 𝑓 of 𝑥 will also be
differentiable for all real values of 𝑥. Next, 𝑔 of 𝑥 is just equal to
𝑥. This is a linear function, so it’s
differentiable for all real values of 𝑥.
Next, we need to show that 𝑔 prime
of 𝑥 is not equal to zero around 𝑥 is equal to zero. The easiest way to do this is to
find an expression for 𝑔 prime of 𝑥. In this case, that’s the derivative
of 𝑥 with respect to 𝑥 which we know is equal to one. And we can see this isn’t equal to
zero for any value of 𝑥. Next, we need to show both the
limit as 𝑥 approaches zero of 𝑓 of 𝑥 and the limit as 𝑥 approaches zero of 𝑔 of
𝑥 are equal to zero. We could do this directly. We could use direct substitution to
evaluate both of these limits. After all, we already know 𝑓 and
𝑔 are continuous when 𝑥 is equal to zero.
However, in fact, we’ve already
done this. Remember what happened when we
tried to evaluate our limit by using direct substitution. We substituted 𝑥 is equal to zero
into our function and we got the indeterminate form zero divided by zero. But think what this means when we
consider just our numerator. We substituted 𝑥 is equal to zero
into our numerator and we got zero. But our numerator is the function
𝑓 of 𝑥. So the limit as 𝑥 approaches zero
of 𝑓 of 𝑥 is equal to zero. The same is true if we just look at
our denominator. We show the limit as 𝑥 approaches
zero of 𝑔 of 𝑥 is equal to zero.
We’ve now shown all our
prerequisites for L’Hôpital’s rule are true. This means we can attempt to
evaluate our limit by evaluating the limit as 𝑥 approaches zero of 𝑓 prime of 𝑥
divided by 𝑔 prime of 𝑥. But before we do this, we see we’re
going to need to find an expression for 𝑓 prime of 𝑥. We know 𝑓 prime of 𝑥 would be
equal to the derivative of five 𝑒 to the power of 19𝑥 minus five 𝑒 to the power
of 10𝑥 with respect to 𝑥. And we can evaluate this derivative
term by term by recalling our derivative results for exponential functions.
For any real constant 𝑏, the
derivative of 𝑒 to the power of 𝑏𝑥 with respect to 𝑥 is equal to 𝑏 times 𝑒 to
the power of 𝑏𝑥. Applying this to differentiate our
first term, we get five times 19𝑒 to the power of 19𝑥. And we can simplify this since five
times 19 is equal to 95. We can do the same with the second
term. We’re subtracting five times 10𝑒
the power of 10𝑥, and five times 10 is equal to 50. So we get 𝑓 prime of 𝑥 is equal
to 95𝑒 to the power of 19𝑥 minus 50𝑒 to the power of 10𝑥.
Now, by applying L’Hôpital’s rule,
we can attempt to evaluate the limit as 𝑥 approaches zero of 𝑓 prime of 𝑥 divided
by 𝑔 prime of 𝑥 as the limit of 𝑥 approaches zero of 95𝑒 to the power of 19𝑥
minus 50𝑒 to the power of 10𝑥 all divided by one. And of course, dividing by one
isn’t going to change the value of our limit. And now we’re just evaluating the
limit as 𝑥 approaches zero of the difference between two exponential functions. This is continuous for all real
values of 𝑥, so we can attempt to evaluate this by direct substitution.
We substitute 𝑥 is equal to zero
into our function, giving us 95𝑒 to the power of 19 times zero minus 50𝑒 to the
power of 10 times zero. Both of our exponents simplify to
give us zero and 𝑒 to the zeroth power is equal to one. So this simplifies to give us 95
minus 50 which we can calculate is equal to 45. And this gives us our final
answer. Therefore, by using L’Hôpital’s
rule, we were able to show the limit as 𝑥 approaches zero of five 𝑒 to the power
of 19𝑥 minus five 𝑒 to the power of 10𝑥 all divided by 𝑥 is equal to 45.
