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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[(\int\limits_{0}^{x}(3t-1)^{10} dt)/3x\]
as x approaches 0 I'm confused because the question has two variables: x and t. How should I approach this?
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Evaluate the limit in the top first.
sorry, I didn't see 3x :)
Use L'Hospital rule
Oh right. L'Hopital's rule is the best choice: \[\frac{d}{dx}\int\limits_0^x(3t-1)^{10}dt=(3x-1)^{10}.\]
The derivative of the bottom is obviously 3. Hence the limit is \(\frac{3(0)-1)^{10}}{3}=\frac{1}{3}\).
This makes sense, but how did you know to use l'hopital's rule?
If we plug x=0, you will get \(\frac{0}{0}\). If you plug x in the top, you get an integral from 0 to 0, which results in a value of 0. I think it's very obvious in the denominator.
oh okay, thanks a ton!

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