next up previous
Next: About this document

MATH 1014.03MW Exam 2 Wednesday, February, 9 2000


Student Number:tex2html_wrap_inline484 No.tex2html_wrap_inline486Marks tex2html_wrap_inline488

Instructions: This exam contains 3 questions each of which has several parts and has a total of 100 marks. Show all of your work.

tex2html_wrap_inline492 (40 Marks) Evaluate each of the following limits.

 (a) tex2html_wrap_inline496
(b) tex2html_wrap_inline500

(a) tex2html_wrap_inline496 (b) tex2html_wrap_inline500

 (c)  tex2html_wrap_inline512.

(c) tex2html_wrap_inline512. (d) tex2html_wrap_inline516.

Solution: (a) tex2html_wrap_inline528.


Hence, tex2html_wrap_inline532.

(3) tex2html_wrap_inline512. =tex2html_wrap_inline538


tex2html_wrap_inline540(20 Marks)Evaluate each improper integral or show that it diverges.

 (a) tex2html_wrap_inline544.
(b) tex2html_wrap_inline548, p>0. ¯

(a) tex2html_wrap_inline544. (b) tex2html_wrap_inline548, p>0.

Solution: (a) x=0 is a singularity. Let tex2html_wrap_inline566. Then

tex2html_wrap_inline568. Hence, we have

This implies tex2html_wrap_inline572.

(b) Let tex2html_wrap_inline576. Then x=1 is a singularity. Let tex2html_wrap_inline566. Then

We consider the following cases:

(i) When 1-p>0, tex2html_wrap_inline588 and tex2html_wrap_inline548 converges.

(ii) When 1-p<0, tex2html_wrap_inline596 and tex2html_wrap_inline548 diverges.

(iii) When p=1, tex2html_wrap_inline604. Hence, tex2html_wrap_inline606. In this case, tex2html_wrap_inline608 diverges.

tex2html_wrap_inline610(40 Marks)Determine convergence or divergence for each of the following series.

 (a) tex2html_wrap_inline614.

(a) tex2html_wrap_inline614. (b) tex2html_wrap_inline618.

 (c) tex2html_wrap_inline630.
(d) tex2html_wrap_inline634

(c) tex2html_wrap_inline630. (d) tex2html_wrap_inline644 (p>0, x>0).

Solution: (a) (Method 1: Bounded Sum Test) Let tex2html_wrap_inline650. Note that tex2html_wrap_inline652 for tex2html_wrap_inline654. Hence, for tex2html_wrap_inline654,

Hence, tex2html_wrap_inline660 is bounded and tex2html_wrap_inline614 converges.

There is another way to show tex2html_wrap_inline660 is bounded. Note that tex2html_wrap_inline666 for tex2html_wrap_inline654. Hence, for tex2html_wrap_inline654,

(Method 2: Ordinary Comparison Test) Let tex2html_wrap_inline674, Then tex2html_wrap_inline676 for tex2html_wrap_inline654. Since tex2html_wrap_inline680 converges, it follows that tex2html_wrap_inline614 converges.

Another way: tex2html_wrap_inline684 for tex2html_wrap_inline654. Since tex2html_wrap_inline688 converges, it follows that tex2html_wrap_inline614 converges.

(Method 3: Ratio Test) tex2html_wrap_inline692. It follows from Ratio Test that the series converges.

(Method 4: Root Test) Noting that tex2html_wrap_inline694 , we have tex2html_wrap_inline696. It follows that tex2html_wrap_inline614 converges.

(b) Since tex2html_wrap_inline702 and tex2html_wrap_inline704 diverges, it follows from Limit Comparison Test that tex2html_wrap_inline706 diverges.

(c) Let tex2html_wrap_inline710. Then tex2html_wrap_inline712 is continuous since tex2html_wrap_inline714 is continuous on tex2html_wrap_inline716 and tex2html_wrap_inline718 for tex2html_wrap_inline720. Also, tex2html_wrap_inline722, so f is decreasing. Moreover,

Hence, tex2html_wrap_inline728 and converges. It follows from Integral Test that tex2html_wrap_inline630 converges.

(d) Let tex2html_wrap_inline734. Then tex2html_wrap_inline736.

Hence, tex2html_wrap_inline740. By Ratio Test, we have

(i) If x<1, tex2html_wrap_inline644 converges.

(ii) If x>1, tex2html_wrap_inline644 diverges.

When x=1, the series becomes tex2html_wrap_inline756. It is known that when p>1, tex2html_wrap_inline756 converges and when tex2html_wrap_inline762, tex2html_wrap_inline756 diverges.

next up previous
Next: About this document

Kunquan Lan
Wed Feb 9 09:35:54 EST 2000