Questions and Answers #7 Limits and Continuity

Questions and Answers Sheet 7 Limits and Continuity Question #1 Evaluate the following limits. √3+8𝑥 2 a) lim 𝑥→∞ 7+11𝑥 √3+8𝑥 2 b) lim 𝑥→−∞ 7+11𝑥 Answer: The given limits √3+8𝑥 2 a) lim 𝑥→∞ 7+11𝑥 √3+8𝑥 2 b) lim 𝑥→−∞ 7+11𝑥 √3+8𝑥 2 a) lim 𝑥→∞ 7+11𝑥 3 √𝑥 2 ( 2 +8) 𝑥 = lim 7 𝑥→∞ 𝑥(𝑥+11) 3 𝑥√ 2 +8 𝑥 = lim 7 𝑥→∞ 𝑥(𝑥+11) 3 √ 2 +8 𝑥 = lim = = = 7 𝑥→∞ 𝑥+11 3 +8 ∞ 7 +112 ∞ 0+8 0+11 8 11 b) here lim → − So there will be no effect answer remains 8 11 Question #2 Use the precise definition of a limit to prove the following limits. lim 𝑥→3 𝑥 2 −7𝑥+12 𝑥−3 = −1 Answer: To prove lim 𝑥→3 𝑥 2 −7𝑥+12 𝑥−3 = −1 Considering LHS lim 𝑥→3 lim 𝑥→3 lim 𝑥→3 lim 𝑥→3 𝑥 2 −7𝑥+12 𝑥−3 𝑥 2 −7𝑥+12 𝑥−3 𝑥 2 −7𝑥+12 𝑥−3 𝑥 2 −7𝑥+12 𝑥−3 = lim 𝑥→3 = lim 𝑥→3 = lim 𝑥→3 𝑥 2 −3𝑥−4𝑥+12 𝑥−3 𝑥(𝑥−3)−4(𝑥−3) 𝑥−3 (𝑥−3)(𝑥−4) 𝑥−3 = lim(𝑥 − 4) 𝑥→3 Now putting limit x=3, in equation lim 𝑥→3 𝑥 2 −7𝑥+12 𝑥−3 = (3 − 4) 𝑥 2 −7𝑥+12 lim 𝑥−3 𝑥→3 = −1 Hence proved Question #3 Find the limits. a) lim −𝑥 𝑥→∞√𝑥 2 +5𝑥 𝜋𝑥 b) lim sin ( 𝑥→∞ 2−3𝑥 ) Answer: Given: a) lim −𝑥 𝑥→∞√𝑥 2 +5𝑥 𝜋𝑥 b) lim sin ( 𝑥→∞ 2−3𝑥 ) Calculation: Multiple and divide the conjugate of √𝑥 2 + 5𝑥 − 𝑥 √(𝑥 2 + 5𝑥)2 − 𝑥 2 √𝑥 2 + 5𝑥 − 𝑥 √𝑥 2 + 5𝑥 + 𝑥 = √𝑥 2 + 5𝑥 + 𝑥 √𝑥 2 + 5𝑥 + 𝑥 5𝑥 = 2 √𝑥 +5𝑥+𝑥 Divided by highest denominator power, 𝑥 = 5 lim 5 𝑥→∞ √1+ +1 𝑥 =5 1 2 b) lim sin ( 𝑥→∞ 𝜋𝑥 2−3𝑥 ) = lim sin ( 𝑥→∞ 𝜋𝑥 2 𝑥 𝑥( −3) ) 𝜋 lim sin (2 ) 𝑥→∞ 𝑥 −3 𝜋 = sin (− ) 3 =− √3 2 Answer: a) lim −𝑥 = 𝑥→∞√𝑥 2 +5𝑥 𝜋𝑥 b) lim sin ( 𝑥→∞ 2−3𝑥 )= 5 2 3 2 Question #4 The process by which we determine limits of rational functions applies equally well to ratios containing non integrer or negative powers of x: Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits lim 4−3𝑥 3 𝑥→−∞ √𝑥 6 +9 Answer: To find the limit lim 4−3𝑥 3 𝑥→−∞ √𝑥 6 +9 On calculating lim 4−3𝑥 3 𝑥→−∞ √𝑥 6 +9 1 (4−3𝑥 3 ) 𝑥3 1 𝑥→−∞ 3 √𝑥 6 +9 𝑥 = lim 4 −3 𝑥3 𝑥→−∞ −√1+ 9 𝑥6 = lim = 0−3 −√1+0 =3 Question #5 The process by which we determine limits of rational functions applies equally well to ratios containing nonintegrer or negative powers of x: Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits lim √𝑥 2 +1 𝑥→−∞ 𝑥+1 Answer: To evaluate: lim √𝑥 2 +1 𝑥→∞ 𝑥+1 Solution: lim √𝑥 2 +1 𝑥→∞ 𝑥+1 On simplifying further we get: lim √𝑥 2 +1 𝑥→∞ 𝑥+1 1 = lim √𝑥 2 (1+ 2 ) 𝑥 1 𝑥 𝑥(1+ ) 𝑥→∞ 1 = lim |𝑥|√(1+ 2 ) 𝑥 1 𝑥 𝑥(1+ ) 𝑥→∞ 1 = lim = = 𝑥√(1+ 2 ) 𝑥 1 𝑥→∞ 𝑥(1+𝑥) √1+0 1+0 1 1 =1 √𝑥 2 +1 ⇒ lim 𝑥→∞ 𝑥+1 =1 Result: lim √𝑥 2 +1 𝑥→∞ 𝑥+1 =1 Question #6 1 Evaluate lim+ − 𝑥→0 𝑥 1 𝑒 𝑥 −1 Answer: L-Hospitals Rule: If lim 𝑓(𝑎) 𝑥→𝑎 𝑔(𝑎) lim 0 = , then 0 𝑑 𝑓(𝑎) 𝑥→𝑎 𝑔(𝑎) = lim 𝑑𝑥 𝑑 𝑥→𝑎 𝑑𝑥 [𝑓(𝑎)] [𝑔(𝑎)] 1 To evaluate: lim+ − = lim+ 𝑒 𝑥 −1−𝑥 𝑥→0 𝑥 = lim 1 𝑒 𝑥 −1 𝑒 𝑥 −𝑥−1 𝑥→0 𝑥(𝑒 𝑥 −1) 𝑥→0+ 𝑥𝑒 𝑥 −𝑥 𝑥 −𝑥−1 𝑒 0 As 𝑥 → 0+ , lim 𝑥 = indeterminate 0 𝑥→0 𝑥𝑒 −𝑥 𝑒 𝑥 −𝑥−1 𝑒 0 −0−1 1−1 0 + As 𝑥 → 0 , lim+ 𝑥→0 𝑥𝑒 𝑥 −𝑥 = 0×𝑒 0 −0 = 0 = 0 which is intederminate form we can use L-Hospitals Rule: ∴ lim+ 𝑒 𝑥 −𝑥−1 𝑥→0 𝑥𝑒 𝑥 −𝑥 = lim+ 𝑑 𝑥→0 𝑒 𝑥 −1−0 𝑥→0 𝑥𝑒 𝑥 +𝑒 𝑥 −1 𝑑𝑥 = lim+ 𝑥→0 𝑒 𝑥 −0 = lim+ 𝑥→0 𝑥𝑒 𝑥 +𝑒 𝑥 +𝑒 𝑥 −0 𝑒𝑥 = lim+ = = (𝑒 𝑥 −𝑥−1) = lim+ 𝑑𝑥𝑑 𝑥→0 𝑥𝑒 𝑥 +2𝑒 𝑥 1 = (𝑥𝑒 𝑥 −𝑥) 𝑑 𝑥 (𝑒 −1) 𝑑𝑥 𝑑 (𝑥𝑒 𝑥 +𝑒 𝑥 −1) 𝑑𝑥 𝑒0 0×𝑒 0 +2×𝑒 0 0×1+2×1 1 1 0+2 = 2 1 So, lim+ − 𝑥→0 𝑥 1 𝑒 𝑥 −1 = 1 2 Question #7 Evaluate the following limits. If you use l'Hospital's Rule, be sure to indicate when you are using it, and why it applies. a) lim (3 ⋅ 21−𝑥 + 𝑥 2 ⋅ 21−𝑥 ) 𝑥→∞ b) lim+ (1 + 5𝑥)2/𝑥 𝑥→0 Answer: Evaluate limits: lim (3 ⋅ 21−𝑥 + 𝑥 2 ⋅ 21−𝑥 ) 𝑥→∞ lim [𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥) 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 lim (3 ⋅ 21−𝑥 + 𝑥 2 ⋅ 21−𝑥 ) = lim (3 ⋅ 21−𝑥 ) + lim (𝑥 2 ⋅ 2−𝑥 ) 𝑥→∞ 𝑥→∞ 1 = 3 ⋅ lim (21−𝑥 ) + 2 ⋅ lim (𝑥 2 𝑥→∞ Take 3 ⋅ lim 𝑥→∞ (21−𝑥 ) 𝑥→∞ 2𝑥 𝑥→∞ ) Apply exponent rule 𝑎 𝑥 = 𝑒 ln(𝑎 = 3 ⋅ lim (𝑒 (1−𝑥) ln(2) ) 𝑥) = 𝑒 𝑥 ln(𝑎) 𝑥→∞ = 3 ⋅ lim (21−𝑥 ) 𝑥→∞ = 3 ⋅ lim (𝑒 𝑥 ln(2) ) 𝑥→−∞ Apply the Limit Chain Rule: 𝑔(𝑥) = 𝑥 ln(2) , 𝑓(𝑢) = 𝑒 𝑢 = −∞ ⋅ ln(2) = −∞ 3 ⋅ lim (𝑒 𝑥 ln(2) ) = 3 ⋅ 𝑒 −∞ 𝑥→−∞ = 3⋅0 =0 Take 2 ⋅ lim (𝑥 2 𝑥→∞ 1 2𝑥 ) If ∑ 𝑎𝑛 converges, then lim (𝑎𝑛 ) = 0 𝑛→∞ Apply ratio test and check weather series is convergent or divergent. 𝑎 If ∣ 𝑛+1 ∣≤ 𝑞 eventually for some 0 1 eventually then ∑∞ 𝑛=1 𝑎𝑛 diverges. lim ( ) = 0 𝑥→∞ 2𝑥 = 2⋅0 =0 lim (3 ⋅ 21−𝑥 + 𝑥 2 ⋅ 21−𝑥 ) = 3 ⋅ lim (21−𝑥 ) + 2 ⋅ lim (𝑥 2 𝑥→∞ =0+0 𝑥→∞ 𝑥→∞ 1 2𝑥 ) =0 lim (3 ⋅ 21−𝑥 + 𝑥 2 ⋅ 21−𝑥 ) 𝑥→∞ Question #8 Evaluate the following limits. √7+9𝑥 2 a) lim 𝑥→∞ 11+7𝑥 √7+9𝑥 2 b) lim 𝑥→−∞ 11+7𝑥 Answer: lim √7+9𝑥 2 𝑥→∞ 11+7𝑥 2 taking 𝑥 lim common from numerator 7 𝑥√ 2 +9 𝑥 𝑥→∞ 11+7𝑥 taking x common from denominator 7 lim 𝑥(√ 2 +9) 𝑥 11 𝑥→∞ 𝑥( 𝑥 +7) 7 = lim √ 2 +9 𝑥 11 𝑥→∞ 𝑥 +7 as 𝑥 → ∞, = 7 𝑥2 1 𝑥2 → 0 and So answer is b) lim 1 → 0 and 11 0+7 𝑥→−∞ 11+7𝑥 →0 →0 𝑥 √0+9 √7+9𝑥 2 𝑥 = 3 7 7 = lim 𝑥√ 2 +9 𝑥 11 𝑥→−∞ 𝑥( 𝑥 +7) 7 +9 2 x lim x →− 11 (( + 7) x as 𝑥 → −∞, 1 1 𝑥2 →0⇒ 𝑥 → −∞; → 0 ⇒ = √0+9 0+7 = 𝑥 3 11 𝑥 7 𝑥2 →0 →0 7 Question #9 Evaluate the following limits. a) lim (−40𝑥 2 − 36𝑥 3 ) 𝑥→∞ b) lim (−40𝑥 2 − 36𝑥 3 ) 𝑥→−∞ Answer: Given a) lim (−40𝑥 2 − 36𝑥 3 ) 𝑥→∞ replacing limit = −40(∞)2 − 36(∞)3 as we know power of infinite is infinite = −40(∞) − 36(∞) multiplication with infinite of any number is infinite = −∞ − ∞ Answer: −∞ Given b) lim (−40𝑥 2 − 36𝑥 3 ) 𝑥→−∞ replacing limit = −40(−∞)2 − 36(−∞)3 as we know power of infinite is infinite = −40(∞) − 36(∞) Multiplication with infinite of any number is infinite = −∞ + ∞ =undefined Question #10 Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits lim ( 𝑥 2 +𝑥−1 8𝑥 2 −3 𝑥→−∞ 1/3 ) Answer: Given, 1 lim ( 𝑥→−∞ 𝑥 2 +𝑥−1 3 ) 8𝑥 2 −3 Concept used: (𝑎𝑚 𝑏)𝑛 = 𝑎𝑚𝑛 (𝑏)𝑛 Apply the above concept, we get 1 lim ( 𝑥→−∞ 2 𝑥 2 +𝑥−1 3 𝑥3 ) = lim ( 2) ( 8𝑥 2 −3 𝑥→−∞ 𝑥3 1 1 1 1+ − 2 3 𝑥 𝑥 3 8− 2 𝑥 ) 1 = lim (1) ( 1 1 1+ − 2 3 𝑥 𝑥 3 8− 2 𝑥 𝑥→−∞ 1 (−∞)2 3 8− (−∞)2 1 1−0−0 3 1 3 1 =( =( 1+−∞− 8−0 ) ) ) 1 1 3 =( ) 8 1 1 3 2 = ( 3) = 1 2 Question #11 Find the limit if it exists. If it does hot exíst explain why Ih the case of ihfinite limits, malk suve that you check both the left and right limit at the point in the question. 1 lim (𝑥 2 −1)𝑒 𝑥−1 (𝑥−1) 𝑥→1 Answer: 1 lim (𝑥 2 −1)𝑒 𝑥−1 𝑥→1 (𝑥−1) 1 𝐿. 𝐻. 𝐿 = lim = lim ℎ→0 ((1−ℎ)2 −1)𝑒 𝑥−ℎ−𝑥 (𝑥−ℎ−𝑥) ℎ→0 𝑥+ℎ2 −2ℎ−𝑥 1 −ℎ 𝑒ℎ 1 = lim − (ℎ − 2)𝑒 −ℎ ℎ→0 Taking limit: L.H.L=0 1 𝑅. 𝐻. 𝐿. = lim = lim ((1+ℎ)2 −1)𝑒 𝑥+ℎ−𝑥 ℎ→0 𝑥+ℎ2 +2ℎ−𝑥 ℎ ℎ→0 = lim (ℎ + 2) 𝑒 𝑥+ℎ−𝑥 𝑒 1 ℎ 1 ℎ ℎ→0 Taking limit 𝑅. 𝐻. 𝐿 = ∞ (does not exist finity) ⇒ 𝐿. 𝐻. 𝐿 ≠ 𝑅. 𝐻. 𝐿 1 ⇒ lim (𝑥 2 −1)𝑒 𝑥−1 (𝑥−1) 𝑥→1 does not exist Question #12 Evaluate each limit and justify your answer. lim ( 𝑥+5 4 𝑥→1 𝑥+2 ) Answer: The given limit is: lim ( 𝑥+5 4 𝑥→1 𝑥+2 ) When substituting the direct limit, this limit will not undefined. Therefore, to evaluate this take the limit directly. That is, lim ( 𝑥+5 4 ) =( 1+5 4 𝑥→1 𝑥+2 𝑥+5 4 1+2 6 4 𝑥→1 𝑥+2 𝑥+5 4 3 lim ( ) =( ) lim ( ) = (2)4 lim ( ) = 16 ) 𝑥→1 𝑥+2 𝑥+5 4 𝑥→1 𝑥+2 Answer: lim ( 𝑥+5 4 𝑥→1 𝑥+2 ) = 16 Question #13 Find the limits lim 𝑥→∞√ 8𝑥2−3 2𝑥2+𝑥 Answer: The degree of both numerator and denominator is=2 So we divide numerator and denominator by 𝑥 2 Calculation: lim 𝑥→∞√ 8𝑥2−3 2𝑥2+𝑥 (8 x 2 − 3) x2 = lim x → (2 x 2 + x) x2 = lim 𝑥→∞√ =√ =√ 3 (8− 2 ) 𝑥 1 (2+𝑥) (8 − 0) (2 + 0) 8 2 = √4 =2 Answer: 2 Question #14 Limits of composite functions Evaluate each limit and justify your answer x3 − 2 x 2 − 8 x lim x →4 x−4 Answer: Given limit is: 𝐿= lim 𝑥→4√ 𝑥3 −2𝑥2 −8𝑥 𝑥−4 Then we get, x( x2 − 2 x − 8) x−4 L = lim x →4 𝐿= lim 𝑥→4√ 𝑥(𝑥−4)(𝑥+2) (𝑥−4) 𝐿 = √4(4 + 2) 𝐿 = √24 𝐿 = 2√6 Hence the value of the given limit is 2√6 Question #15 Determine which of the following limits exist, and find the limits which do exists. a) b) lim 𝑥 3 +𝑦 3 (𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2 𝑥 2 +4𝑥𝑦 2 +4𝑦 4 lim 𝑥 2 +4𝑦 4 (𝑥,𝑦)→(0,0) Answer: Consider the provided question, a) Given that, lim 𝑥 3 +𝑦 3 (𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2 Put y=mx lim 𝑥 3 +𝑦 3 (𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2 = lim = lim 33 +(𝑚𝑥)3 (𝑥,𝑚𝑥)→(0,0) 𝑥 2 +(𝑚𝑥)2 𝑥 3 (1+𝑚3 ) 𝑥→0 𝑥 2 (1+𝑚2 ) 𝑥(1+𝑚3 ) = lim 𝑥→0 (1+𝑚2 ) = 0 (Limits is 0 for all values of m) Hence, Limit exist and lim 𝑥 3 +𝑦 3 (𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2 =0 Question #16 Limits of composite functions Evaluate each limit and justify your answer 𝑡−4 lim tan √𝑡−2 𝑥→4 Answer: Given limit is: 𝑡−4 𝐿 = lim tan √𝑡−2 𝑥→4 Then we get, 𝐿 = lim [tan (√𝑡+2)(√𝑡−2) ] √𝑡−2 𝑡→4 𝐿 = lim[tan(√𝑡 + 2)] 𝑡→4 𝐿 = tan(√4 + 2) 𝐿 = tan(2 + 2) 𝐿 = tan 4 Hence the value of this limit is tan 4 Question #17 Find the limit. If limit does not exist state it. In each part show your calculations and circle the answer lim 5𝑥 2 +7 𝑥→∞ 3𝑥 2 −𝑥 Answer: lim 5𝑥 2 +7 7 = lim 𝑥 2 (5+ 2 ) 𝑥 1 𝑥→∞ 3𝑥 2 −𝑥 𝑥→∞ 𝑥 2 (3−𝑥) 7 lim 5+ lim 2 𝑥→∞ 𝑥→∞𝑥 1 lim 3− lim 𝑥→∞ 𝑥→∞𝑥 7 = lim 5+ 2 𝑥 1 𝑥→∞ 3−𝑥 = 5+0 5 = 3−0 3 5𝑥 2 +7 5 = lim 2 = = 𝑥→∞ 3𝑥 −𝑥 3 Question #18 Evaluate the given limits. a) lim(2 sin 𝑥 + 𝑒 𝑥 ) 𝑥→0 𝑥−2 b) lim 𝑥→2 𝑥 2 −4 Answer: Solution: Here we find the limits of follow: a) lim(2 sin 𝑥 + 𝑒 𝑥 ) 𝑥→0 = lim 2 sin 𝑥 + lim 𝑒 𝑥 𝑥→0 𝑥→0 = 2 lim sin 𝑥 + lim 𝑒 𝑥 𝑥→0 𝑥→0 = 2⋅0+1 =1 Hence lim(2 sin 𝑥 + 𝑒 𝑥 ) = 1 b) lim 𝑥→0 𝑥−2 𝑥→2 𝑥 2 −4 𝑥−2 = lim (𝑥−2)(𝑥+2) 𝑥→2 = lim 1 𝑥→2 𝑥+2 = 1 4 Hence we get 𝑥−2 1 lim 2 = 𝑥→2 𝑥 −4 4 Question #19 Compute the limits. Justify your answer. 16𝑥 3 −19𝑥 2 +7𝑥+11 a) lim 𝑥→∞ 5𝑥 2 −10𝑥 3 −19𝑥+100 𝑥 5/7 b) lim 𝑥→∞ 17𝑥+29 Answer: The limit of given function is computed directly by using some results a) lim 16𝑥 3 −19𝑥 2 +7𝑥+11 5𝑥 2 −10𝑥 3 −19𝑥+100 𝑥→∞ 19 7 11 𝑥 3 (16− + 2 + 3 ) = lim 16𝑥 3 −19𝑥 2 +7𝑥+11 𝑥→∞ −10𝑥 3 +6𝑥 2 −19𝑥+100 𝑥 𝑥 𝑥 5 19 100 = lim 𝑥→∞ 𝑥 3 (−10+𝑥− 2 + 3 ) 𝑥 𝑥 19 7 11 + 2 + 3) x x x = lim x → 5 19 100 −10 + − 2 + 3 ) x x x (16 − = 16 −10 −8 = 5 b) lim 𝑥5/7 ) 𝑥 17𝑥+29 𝑥→∞ 𝑥( 𝑥 ) 𝑥( 𝑥 5/7 𝑥→∞ (17𝑥+29) = lim 𝑥5/7 ) 𝑥 17𝑥+29 𝑥→∞ ( 𝑥 ) ( = lim 1 ( 2/7 ) 𝑥 ⇒ lim 29 𝑥→∞ (17+ 𝑥 ) 𝑥 5/7 ⇒ lim 𝑥→∞ 17𝑥+29 = 0 17+0 =0 =0 Question #20 Determine the following limits, using ∞ or −∞ when appropriate, or state that they do not exist. lim 𝑥→0 𝑥 4 +11𝑥 3 𝑥3 Answer: lim 𝑥→0 𝑥 4 +11𝑥 3 𝑥3 Factor: lim 𝑥→0 𝑥 4 +11𝑥 3 𝑥3 = lim 𝑥 3 (𝑥+11) 𝑥3 𝑥→0 Cancel the common term: lim 𝑥→0 𝑥 3 (𝑥+11) 𝑥3 = lim(𝑥 + 11) 𝑥→0 Substitute the variable with the value: lim(𝑥 + 11) = (11) 𝑥→0 Therefore, lim 𝑥→0 𝑥 4 +11𝑥 3 𝑥3 = 11