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Using L'Hopital to Evaluate Limits

L'Hopital's Rule is a method of differentiation to solve indeterminant limits. Indeterminant limits are limits of functions where both the function in the numerator and the function in the denominator are approaching 0 or positive or negative infinity. It is not clear what the limit of indeterminant forms are, but when applying L'Hopital's Rule, indeterminant limits can be made easier to evaluate.

Evaluate the following limits:

(1)

(2)

These limits are indeterminant because the quotient on the left is 00 when x = 3, and the limit on the right is -∞ when x approaches infinity. We cannot simply plug in the approaching value for x to find the limit. Luckily, there are different methods we can use.

(1) For the first limit, we could factor out an (x-3)

It is easy to see that when x is 3, the limit is 6.

(2) For the second limit, we can factor out an x2

Knowing that the limit of any number over infinity is 0, we can plug 0 into the limit and simplify to 6-5.

(3) But what about this limit?

We cannot factor anything out, so how to we evaluate it?

We can see that when x approaches 0, both the numerator and denominator approach 0. Because the quotient will be 00, it is not clear what the limit will be. In the limits page, we evaluated this limit by looking at the graph of the function's behavior as it approached 0 from the left and the right. Using L'Hopital's rule, we can now evaluate the limit in a determinant form. Since both the numerator and the denominator go to 0 and both functions have a deritive, we can apply L'Hopital's rule.

L'Hopital's Rule states that for functions f(x) and g(x):



L'Hopital's Rule

Let's use L'Hopital's rule for our limit.

Differentiating both the numerator and the denominator will simplify the quotient and make evaluating the limit easier.

Taking the derivative of the numerator and denominator, the limit is easier to see. We know that the cos(0) is 1, so the limit as x approaches 0 is 1. To check, we can graph both functions and see that they both converge to y = 1 as x approaches 0.

Let's use L'Hopital's rule on our first two limits to see if it works.

(1) and (2) Evaluate the following limits:

(1)

We take the derivative and plug in 3 for x to get our limit.

(2)

We take the derivative twice and simplify. After the first derivative, the quotient is still -∞, so we can apply L'Hopital's rule again and take the derivative. Simplifying, we get 6-5.

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