## Calculus Practice Exams

Problem 1. Find the limit Answer. Note that we have an indeterminate form . You may want to try Hopital rule and see how difficult the calculations become except if you use the substitution . Here we will use Taylor Polynomials to find this limit. We have This gives Clearly the limit is Problem 2. Find the limit Answer. Note that we have an indeterminate form . Since we a have a square root in the function, we will use the conjugate forms to get rid of it. We have Since (note that |x| = -x because and therefore x < 0), hence It is now obvious that Problem 3. Find Answer. Note that we have an indeterminate form . In order to find the limit, we will use the formula We can assume that . In this case, we have Since we have to concentrate on the term . We have Since we get Therefore we have Problem 4. Determine the convergence or divergence of

(a) Answer. The bad points are , , -1.
At , we have when . Recall that when if and only if Using the limit test and the p-test, we know that are both convergent. Thereofre are both convergent.
At -1, we have when . Again using the limit test and the p-test, we know that is divergent. Therefore the improper integral is divergent.

(b) Answer. The bad points are 0 and .
At , we have when . Using the limit test and the p-test, we know that is divergent. Therefore is divergent. So the improper integral is divergent. We suggest to consider the other bad points too though there is no need for that. But you never know, your previous conclusion may be wrong!!!!
At 0, we have when . Using the limit test and the p-test, we know that is divergent. Therefore is divergent. So again we get the divergence of the hole improper integral.

(c) Answer. The bad points are 1 and .
At 1, we have when . Using the limit test and the p-test, we know that is convergent.

At , we have when . And since when and the improper integral is divergent, then using the limit test and the p-test, we have is divergent. We deduce then that the improper integral is divergent.

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