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Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \displaystyle \lim_{x\to 3} \frac{x - 3}{x^2 - 9} $

$\frac{1}{6}$

02:41

Mengsha Y.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

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So this problem here is going to have two different methods. We know that our graph is going to be X -3 divided by x squared minus nine. But remember the x squared minus nine can be written as X plus three times x minus three. Based on this, we see that the X -3 is will cancel just giving us one over X plus three. So based on this, we see that when we just plug in a three we'll get 1/6 as our limit. So 1/6 is our limit. Um That will be the answer. We can also use little cows rule though, because as we saw before, if we plugged in right here, if we plugged in a three, we would get 0/0, which would be the indeterminant form based on that. We can use locales rule to give us Um when we took the driver of this, we get one and we take the derivative of this, we would get to X based on that when we plug in a three, then once again it'll be 1/2 times three. So that would give us 1/6 as the final answer. So you see in two cases we'll get 1/6. Oftentimes there won't be that first option, but it is nice to be able to see it because it's easier not to have to take the derivative if you don't have to, it's easier to recognize that we can just factor and then um cancel the numerator and denominator and then evaluate the limits. Okay?

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