MATH 41 Midterm: Stanford MATH 41 10exam1sol

31 views12 pages

Document Summary

Solutions to math 41 first exam october 12, 2010: (13 points) find each of the following limits, with justi cation. If the limit does not exist, explain why. We may now substitute in x = 5, which yields the limit: e5. = lim x 4 x 2 x 3(cid:19) (by splitting up terms) x2 x2 + 1 x 2 x 3 x4(cid:19) (multiply each term by 1) x 4 x4 x2 x! (simplifying). Hence, our desired limit is lim x x. 1 x(cid:19) (4 points) let f (x) = cos(cid:0)sin 1 x(cid:1); we may then rewrite. 1 cos( ) 1 for all , As x 0+, we may assume x is positive and therefore that x exists; hence, we multiply all three sides to obtain. 1 f (x) 1 for all x . X x f (x) x . x 0+ x and. Since the two limits both exist and equal 0, the squeeze theorem says that lim x 0+

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers

Related textbook solutions

Related Documents

Related Questions