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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.$\displaystyle \lim_{x\to 1} \frac{x^3 - 2x^2 + 1}{x^3 - 1}$

## This limit has the form $\frac{0}{0} . \quad \lim _{x \rightarrow 1} \frac{x^{3}-2 x^{2}+1}{x^{3}-1} \stackrel{\Perp}{=} \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x}{3 x^{2}}=-\frac{1}{3}$Note: Alternatively, we could factor and simplify.

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in this case we could use another method other than the hotel's rule. What we see is that it would give us actually in this case we would not be able to just give us the indeterminant form when we plug in X equals one. So because of that we want to look at the limit or the derivative of the numerator and the denominator now, so when we take the derivative, what we end up getting is three X squared minus for X. Provided by three x square. Now, when we plug in one, what we end up getting as a result Is going to be 3 -4 divided by free. So our final answer is going to be a negative one third. As a result, sometimes we can simplify it. So that way we don't have to use the hotel's rule, but in this case we have to use the hotel's role. Um and as long as we know how to take the derivative properly, it's fairly simple.

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