L'Hospital's Rule

Find the limit:

1

Because both the sine function sin x and exponential function e^x are continuous, we can move the limit symbol through the exponential and sine functions.

In other words, because e^x and sin x are both continuous at x = 0, we can simply plug in x = 0 to find the limit.

2

Applying l'Hospital's Rule gives a rational expression that can be evaluated.

3

The cos x terms divide out (because they are not zero near x = 0), and we can evaluate the limit.

4

Thus, the limit we seek is 1.

5

Some of you may be tempted to think that the value of 1 is the y intercept in the opening figure, but that is incorrect. Our rational function is not defined at x = 0 (its graph is shown at left).

In fact, you may now recognize that this rational function represents the derivative of e^sin(x) at x = 0. Thus, the tangent line to e^sin(x) at x = 0 has slope 1.

WAIT! Weren't we warned in the Introduction not to use l'Hospital's Rule to evaluate derivative expressions?

6

That's true if the derivative you use in the evaluation is the derivative you are trying to evaluate.

Here, the derivatives of sin x and e^x are derived independently of e^{sin x}, which also requires the Chain Rule (which has also been derived independently of our use of l'Hospital's rule).

Show that,

1

Let .

First, let's let r = 1 (rate of 100%) and show,

2

Define some functions of n to make the notation easier.

Note that because ln(1) = 0,

.

By using the reciprocal of n, the limit is converted to the indeterminate form 0/0.

3

The limit on L(n) as n → 0 is the indeterminate form ∞ ⋅ 0, since  and ln(1) = 0.

4

Rewrite the expression using the reciprocal of n.

The limit now has the indeterminate form 0/0.

5

The limit in n (a discrete variable) is the same as the limit in x (a continuous variable).

6

Apply l'Hospital's Rule by taking the derivatives of the numerator and denominator.

7

We can divide out the terms    because they are not zero.

8

The limit is now well-defined because , and the result is 1.

9

Knowing the limit on L(x) can be used to find the limit on A(x).

10

Because , we have that

.

Rewrite, using the properties of logarithms and exponential functions.

11

Using the substitution rule for limits of continuous functions, we can determine the limit for A(x).

12

Applying our result on the limit of L(x) gives our result.

13

Show  .

We now return to our original limit problem.

14

Rewrite the limit expression using the law of exponents.

15

Because the limit does not depend on r, we can take the limit inside the exponential expression.

16

Let .

17

The , thus

Use our previous result to take the limit.

Although p₄(x) is greater at first, the exponential function eventually becomes greater for x > 9.

Show that,

for any N^th‑order polynomial p_N(x).

That is, exponential growth is eventually faster than any polynomial growth.

1

Define p_N(x) as an arbitrary N^th‑degree polynomial, that is, a_N ≠ 0.

2

Continuing …

Examine the derivatives of p_N(x).

Note that the N^th derivative of p_N(x) is a constant, whereas the lower-order derivatives depend on x (because a_N ≠ 0, the leading term of the derivative will always depend on a power of x).

3

Now let's consider our limit and apply l'Hospital's Rule.

4

A repeated application of l'Hospital's Rule reduces the degree of the numerator, but we still have an indeterminate form.

5

for

Because , the denominator term will not change and the numerator term, for the derivative order k < N, will depend on x and thus will go to infinity in the limit.

6

But the N^th derivative (application of l'Hospital's Rule) results in a constant nonzero numerator term (thus, we can no longer apply l'Hospital's Rule).

7

 for
that is,

Although N! can be extremely large (try calculating 70! on your calculator), eventually, e^x is larger.

8

In fact, using the properties of the logarithm, we can see that even for moderately large x, the exponential is quite large.

9

The table at left shows that for x > 365 plus the ln(a_N), the exponential will start growing faster than a 100th-degree polynomial.

Similar estimates can be calculated for higher-order polynomials, but the exponential will always grow faster eventually.

10

The graph at left is of the ratio

.

The red dot (D) shows where the ratio becomes 1; that is, e^x intersects and overtakes x¹⁰ (at x approximately 35.7715).

Notice the y-axis scale and how the polynomial growth dominates for x < 10 but then starts to fall behind the exponential growth, eventually losing out as the ratio goes to zero.

Find the (asymptotic) limit:

1

The limit is of the indeterminate form ∞/∞, but the graph, shown above left, indicates that the function may level off to an asymptote.

2

Apply l'Hospital's Rule.

3

Simplifying gives a result that is the reciprocal of our original rational expression and still has an indeterminate form. So let's apply l'Hospital's Rule again.

4

Taking the derivatives and simplifying, we see that the result is our original rational expression!

Thus, if we continue to apply l'Hospital's Rule, the results will just keep repeating and will always be indeterminate forms. Thus, l'Hospital's Rule fails in this case!

5

Because the graph does indicate a limit (asymptotic behavior), let's try another method from Calculus I.

Multiply the numerator and denominator by . We may do so because taking the limit at 0 means that x ≠ 0, so  is defined.

6

In order to move the x inside the square root, it is rewritten as . Note that, in fact, , but because in the limit x → ∞, we can assume x > 0 and write |x| = x.

7

Because  approaches 0 as x → ∞, the limit of our rational expression is 1, which is the same as indicated on our graph.

8

The graph at left shows the horizontal asymptote at y = 1 as derived from our limit calculations.

Do not blindly apply l'Hospital's Rule to an indeterminate limit—examine it first to see if the indeterminate form may be solved with standard limit procedures.

A graph can help, but remember that a graph is not a proof of the actual limit value.