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Applying L'Hopital's Rule in Simple Cases

Applying L'Hopital's Rule in Simple Cases
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  • 0:06 Quick Review of…
  • 0:37 L'H?pital's Three-Step Plan
  • 0:55 Example #1
  • 2:51 Example #2
  • 4:35 Example #3
  • 6:00 Example #4
  • 7:14 Lesson Summary
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Lesson Transcript
Instructor: Jeff Calareso

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

L'Hôpital's rule may have disputed origins, but in this lesson you will use it for finding the limits of a range of functions, from trigonometric to polynomials and for limits of infinity/infinity and 0/0.

Quick Review of L'Hôpital's Rule

Use LHopitals rule to find the limits of a range of functions
Lhopitals Rule

Do you remember those two bickering mathematicians, L'Hôpital and Bernoulli? Well, let's take a better look at the rule that they fought over. L'Hôpital's rule says that when we're trying to find the limit of something, like f(x)/g(x), and we end up finding 0/0, or infinity/infinity, we can instead find the limit of f`(x)/g`(x). Let's look at a few examples.

L'Hôpital's Three-Step Plan

To do these examples, we're going to follow L'Hôpital's Three-Step Plan:

  1. Check the limit of the top and bottom.
  2. Differentiate the top and bottom.
  3. Calculate the limit of those derivatives we just found.

Example #1

For example, what's the limit of (sin(2x))/x as x goes to zero? So step one, let's try to calculate this without doing any differentiation. Well, sin(2x), as x goes to zero, is like sin(0), which is zero; x will also be zero. So the limit, as x goes to zero, of (sin(2x))/x really is 0/0. This tells me immediately that I need to use L'Hôpital's rule. Step two, I need to differentiate the top and the bottom of this equation. So d/dx of sin(2x) is 2cos(2x); d/dx of x is just 1.

Step three, calculate the limit of the derivatives. This just means I'm going to plug in my derivative for the top and the bottom into L'Hôpital's rule. So the limit, as x goes to zero, of (sin(2x))/x is equal to the limit, as x goes to zero, of (2cos(2x))/1. When I solve this, I see that 2cos(2x) is going to go to 2 as x goes to zero because cos(0) is going to go to 1, and I'm multiplying that by 2. The limit of 1 as x goes to zero is just 1. My limit is 2. So the limit, as x goes to zero, of (sin(2x))/x is just 2.

The solution for example #1
Example of LHopitals

Example #2

Let's look at a second example. What about the limit, as x goes to zero, of (sin(x) - x)/x? Three step plan: First, check the limit of the top and the bottom. Okay, so sin(x) - x; what is the limit of that as x goes to zero? Well, as x goes to zero, sin(x) goes to zero and x is obviously zero. So I have zero on the top. On the bottom, I just have x, and as x goes to zerp, that also goes to zero. So we get 0/0. Time to use L'Hôpital's rule.

Second step: Differentiate the top and the bottom. Well, the derivative of (sin(x) - x) is equal to the derivative of sin(x) - so cos(x) - minus the derivative of x, which is 1. So the derivative of the top is cos(x) - 1. The derivative of the bottom is the derivative of x, which is just 1. Now I'm going to plug those derivatives in to L'Hôpital's rule. So the limit, as x goes to zero, of (sin(x) - x)/x is equal to the limit, as x goes to zero, of (cos(x) - 1)/1. Well, as x goes to zero, the top here goes to zero, because cos(x) will go to 1, and 1 - 1 = 0. And the bottom will go to 1. So this tells us that the limit, as x goes to zero, of (sin(x) - x)/x is zero.

Example #3

The final result in the third example
LHopital Rule Simple Example 3

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