Calculus Practice Exams

Answers.


Problem 1. Find the limit

displaymath234

Answer. Note that we have an indeterminate form tex2html_wrap_inline240 . You may want to try Hopital rule and see how difficult the calculations become except if you use the substitution tex2html_wrap_inline242 . Here we will use Taylor Polynomials to find this limit. We have

displaymath244

This gives

displaymath246

Clearly the limit is

displaymath248

Problem 2. Find the limit

displaymath235

Answer. Note that we have an indeterminate form tex2html_wrap_inline250 . Since we a have a square root in the function, we will use the conjugate forms to get rid of it. We have

displaymath252

Since

displaymath254

(note that |x| = -x because tex2html_wrap_inline258 and therefore x < 0), hence

displaymath262

It is now obvious that

displaymath264

Problem 3. Find

displaymath236

Answer. Note that we have an indeterminate form tex2html_wrap_inline266 . In order to find the limit, we will use the formula

displaymath268

We can assume that tex2html_wrap_inline270 . In this case, we have

displaymath272

Since

displaymath274

we have to concentrate on the term tex2html_wrap_inline276 . We have

displaymath278

Since

displaymath280

we get

displaymath282

Therefore we have

displaymath284

Problem 4. Determine the convergence or divergence of

(a)
tex2html_wrap_inline286

Answer. The bad points are tex2html_wrap_inline288 , tex2html_wrap_inline290 , -1.
At tex2html_wrap_inline294 , we have

displaymath296

when tex2html_wrap_inline298 . Recall that

displaymath300

when tex2html_wrap_inline302 if and only if

displaymath304

Using the limit test and the p-test, we know that

displaymath306

are both convergent. Thereofre

displaymath308

are both convergent.
At -1, we have

displaymath312

when tex2html_wrap_inline314 . Again using the limit test and the p-test, we know that

displaymath316

is divergent. Therefore the improper integral

displaymath318

is divergent.

(b)
tex2html_wrap_inline320

Answer. The bad points are 0 and tex2html_wrap_inline290 .
At tex2html_wrap_inline290 , we have

displaymath326

when tex2html_wrap_inline328 . Using the limit test and the p-test, we know that

displaymath330

is divergent. Therefore

displaymath332

is divergent. So the improper integral

displaymath334

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

displaymath336

when tex2html_wrap_inline338 . Using the limit test and the p-test, we know that

displaymath340

is divergent. Therefore

displaymath342

is divergent. So again we get the divergence of the hole improper integral.

(c)
tex2html_wrap_inline344

Answer. The bad points are 1 and tex2html_wrap_inline346 .
At 1, we have

displaymath348

when tex2html_wrap_inline350 . Using the limit test and the p-test, we know that

displaymath352

is convergent.

At tex2html_wrap_inline346 , we have

displaymath356

when tex2html_wrap_inline358 . And since

displaymath360

when tex2html_wrap_inline358 and the improper integral

displaymath364

is divergent, then using the limit test and the p-test, we have

displaymath366

is divergent. We deduce then that the improper integral

displaymath368

is divergent.


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