L’Hopital’s Rule - Calculus II - Worksheet | MATH 141, Assignments of Calculus

Download L’Hopital’s Rule - Calculus II - Worksheet | MATH 141 and more Assignments Calculus in PDF only on Docsity! MATH 141, Worksheet 6: 7.6, L’Hôpital’s Rule (1) limx→0+ (sinx) (lnx). Solution: This is a 0 · ∞ form. lim x→0+ (sinx) (lnx) = lim x→0+ lnx cscx = lim x→0+ 1 x cscx cotx = lim x→0+ sin2 x x cosx = lim x→0+ 2 cosx sinx 1− x sinx = 0 (2) limx→+∞ 2 x 3x2 Solution For this one, you don’t really need to use L’Hopital. Notice that 3x 2 goes faster to +∞ than 2x. Hence the limit is 0. (3) limx→1 xtan πx 2 . Solution lim x→1 xtan πx 2 = lim x→1 etan πx 2 lnx = elimx→1 tan πx 2 lnx Coming over on the side, lim x→1 tan πx 2 lnx = lim x→1 lnx cot πx2 = lim x→1 1 x −π2 csc2 πx 2 = − 1π 2 = − 2 π Hence, lim x→1 xtan πx 2 = e− 2 π . (4) limx→0+ (ex + 3x) 1/x Solution: lim x→0+ (ex + 3x)1/x = lim x→0+ e 1 x ln(ex+3x) = elimx→0+ 1 x ln(ex+3x) 1 2 So, applying L’hopital, lim x→0+ ln (ex + 3x) x = lim x→0+ ex + 3 ex + 3x = 4 Thus, lim x→0+ (ex + 3x)1/x = e4. (5) limx→0 ( 4 x2 − 21−cosx ) . Solution: lim x→0 ( 4 x2 − 2 1− cosx ) = lim x→0 4− 4 cosx− 2x2 x2 (1− cosx) = lim x→0 4 sinx− 4x 2x− 2x cosx+ x2 sinx = lim x→0 4 cosx− 4 2− 2 cosx+ 4x sinx+ x2 cosx = lim x→0 −4 sinx 2 sinx+ 6x cosx− x2 sinx = lim x→0 −4 cosx 2 cosx+ 6 cosx− 4x sinx− x2 cosx = −2 (6) limx→−∞ x2ex. Solution: lim x→−∞ x2ex = lim x→−∞ x2 e−x = lim x→−∞ 2x −e−x = lim x→−∞ 2 e−x = 0 (7) Review of Exponential and Logarithmic Functions (a) Simplify the following Expression log5(x2 − 1) ln ( x √ x2 − 1 )