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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.$\displaystyle \lim_{x\to 4} \frac{x^2 - 2x - 8}{x - 4}$

## This limit has the form $\frac{0}{0}, \quad \lim _{x \rightarrow 4} \frac{x^{2}-2 x-8}{x-4}=\lim _{x \rightarrow 4} \frac{(x-4)(x+2)}{x-4}=\lim _{x \rightarrow 4}(x+2)=4+2=6$Note: Alternatively, we could apply I'Hospital's Rule.

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So you see this problem here has two options, one with and one without little tells role. First option is if we factor the numerator, we see that. This is the same thing as X plus two Times Active Plus four. And that allows us to cancel with something in the denominator because right down here we have exposed to in the next -4. So based on this we can cancel the X -4. Um and the X minus for giving us X plus tours are line. So when we plug in for um for the limit, we just end up getting six as our answer. But now let's try doing this with the help of cows rule. So basically here, if we plugged in X equals four, we would get For for this would give us 0/0, which is an indeterminate form. So now we have to take the uh We have to use locales role giving us one and the nominator and up in the numerator we would end up getting two X -2. So when we plug in four here, that would be 8 -2, which is 6/1, that's six. So once again we get a final answer of six.

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