Questions and Answers #11 Limits and Continuity

Questions and Answers Sheet 11 Limits and Continuity Question #1 Find the slope of the tangent line to the parabola 𝑦 = 4𝑥 − 𝑥 2 at the point (1,3), i) using Definition 1, ii) using Equation 2 Answer: Definition 2: 𝑚 = lim 𝑓(𝑥)−𝑓(𝑎) 𝑥−𝑎 𝑥⇒𝑎 i) 𝑚 = lim 𝑓(𝑥)−𝑓(1) 𝑥−1 𝑥⇒1 (4𝑥−𝑥 2 )−(4(1)−12 ) = lim 𝑥⇒1 = lim 𝑥⇒1 = lim 𝑥−1 −𝑥 2 +4𝑥−3 𝑥−1 (𝑥−1)(3−𝑥) 𝑥−1 𝑥⇒1 = lim (3 − 𝑥) = 3 − 1 𝑥⇒1 =2 Equation 3: 𝑚 = lim 𝑓(𝑎+ℎ)−𝑓(𝑎) ℎ ℎ⇒0 ii) 𝑚 = lim 𝑓(1+ℎ)−𝑓(1) ℎ ℎ⇒0 [4(1+ℎ)−(1+ℎ)2 ]−[4(1)−12 ] = lim ℎ⇒0 = lim ℎ 2ℎ−ℎ2 ℎ⇒0 ℎ = lim (2 − ℎ) = 2 − 0 ℎ⇒0 =2 Result: m=2 Question #2 How do you find the limit as x approaches infinity of 𝑦 = arctan(𝑒 𝑥 ) ? Answer: 𝜋 As lim 𝑒 𝑥 = +∞ 𝑎𝑛𝑑 lim arctan(𝑥) = . 𝑥⇒+∞ 2 𝑥⇒+∞ Thus, if 𝑥 ⇒ ∞, 𝑒 𝑥 ⇒ ∞ so that, letting 𝑡 = 𝑒 𝑥 , you have: 𝜋 lim arctan(𝑒 𝑥 ) = lim arctan(𝑡) = 𝑥⇒+∞ 2 𝑡⇒+∞ Question #3 Evaluate the limit lim ℎ⇒0 Answer: (𝑥−ℎ)3 −𝑥 3 ℎ lim (𝑥−ℎ)3 −𝑥 3 ℎ 𝑥 3 −3𝑥 2 ℎ+3𝑥ℎ2 −ℎ3 −𝑥 3 ℎ⇒0 𝐿 = lim ℎ ℎ⇒0 −3𝑥 2 ℎ+3𝑥ℎ2 −ℎ3 = lim ℎ ℎ⇒0 = lim − 3𝑥 2 + 3𝑥ℎ − ℎ2 ℎ⇒0 = lim − 3𝑥 2 + 0 + 0 ℎ⇒0 = lim − 3𝑥 2 ℎ⇒0 Question #4 Let 𝑓(𝑥) = (𝑥+8)2 𝑥 2 −64 . Find a) lim 𝑓(𝑥) , b) lim 𝑓(𝑥), and c) lim 𝑓(𝑥). 𝑥⇒−8 𝑥⇒0 𝑥⇒8 Answer: Step 1 a) Plug the given expression for f(x), then factor the denominator and simplify. lim 𝑓(𝑥) 𝑥⇒−8 = lim (𝑥+8)2 𝑥⇒−8 𝑥 2 −64 (𝑥+8)2 = lim 𝑥⇒−8 𝑥 2 −82 (𝑥+8)2 = lim 𝑥⇒−8 (𝑥+8)(𝑥−8) (𝑥+8) = lim 𝑥⇒−8 (𝑥−8) Step 2 Plug x=-8 for the limit = lim = = (𝑥+8) 𝑥⇒−8 (𝑥−8) −8+8 −8−8 0 −16 =0 Question #5 (a) limit as x goes to 0 −  (b) limit as x goes to 5−  tan x 2x −1 ( x − 5) 2 (c) limit as x goes to 0−  cot x 2 Answer: Step 1 a) The limit can be evaluated as. tan 𝑥 0 lim = form use L'Hospitals rule, 𝑥⇒0 2𝑥 lim tan 𝑥 𝑥⇒0 2𝑥 = lim 𝑥⇒0 0 𝑑 tan 𝑥 = lim 𝑑𝑥𝑑 𝑥⇒0 sec2 𝑥 2 𝑑𝑥 2𝑥 = = (12 ) 1 2 (∵ sec 0 = 1) 2 Question #6 ∞ 2 Sequences where ∑∞ 𝑛=𝑘 𝑎𝑛 = ∑𝑛=𝑘 𝑎𝑛 I was recently looking at the series ∑∞ 𝑛=1 = 1 2 sin 𝑛 𝑛 , for which the value quite cleanly comes out to be (𝜋 − 1), which is a rather cool closed form. Answer: Notice that a similar property holds for the related integral ∞ sin 𝑥 ∫0 𝑥 ∞ sin 𝑥 2 𝜋 𝑥 2 𝑑𝑥 = ∫0 ( ) 𝑑𝑥 = This makes me wonder if such sequence has to do something with the property of orthogonality. Two functions f and g are said to be orthogonal with weight 1 ∞ ∫0 𝑓(𝑥)𝑔(𝑥) = 0 If we impose the condition that g(x)=1-f(x) then we are looking at a special case of orthogonality where ∞ ∫0 𝑓(𝑥)(1 − 𝑓(𝑥)) = 0 which is analogous to question since ∑∞ 𝑛=𝑘 𝑎𝑛 (1 − 𝑎𝑛 ) = 0 Question #7 1 Limit of sec ( ) when 𝑥 → ∞ 𝑥 Answer: 1 1 𝑥 𝑛 Assuming sec ( ) instead of sec ( ) , 1 lim sec ( ) = lim+ sec 𝑦 = sec 0 = 1 𝑥→∞ 𝑥 𝑦→0 1 (Taking 𝑦 = ) 𝑥 Question #8 Solving the following limit without L'Hospital's rule: lim sin(𝑥 2 +2)−sin(𝑥+2) 𝑥 𝑥→0 Answer: You can just sum-to-product it from the beginning and use lim lim sin(𝑥 2 +2)−sin(𝑥+2) 𝑥 𝑥→0 = cos(2) lim 𝑥→0 = lim 2 cos[2+𝑥(𝑥+1)/2] sin[𝑥(𝑥−1)/2] 𝑥 𝑥→0 sin[𝑥(𝑥−1)/2] 𝑥→0 = − cos(2) lim 𝑥→0 𝑥/2 sin[𝑥(𝑥−1)/2] 𝑥(𝑥−1)/2 = − cos(2) Question #9 Finding lim ∑𝑛𝑟=1 cot −1 ( 𝑛→∞ Answer: 𝑛2 (𝑟 2 +𝑟+1)+2𝑛+1 𝑛2 +𝑛 ) sin 𝑥 𝑥 =1 Doing some basic manipulation we get the following terms cot −1 ( 1 𝑎 𝑎 𝑎2 (𝑟 2 +𝑟)+1 𝑛 have cot −1 (𝑎(𝑟 2 + 𝑟) + ) = arctan ( arctan(𝑎𝑟) thus sum is ∑∞ 𝑟=1 arctan ( 𝑛+1 𝑛(𝑟 2 +𝑟) 𝑛+1 + 𝑛+1 𝑛 ). Now I let 𝑛 𝑛+1 = 𝑎 so we ) we have y=ar+a,x=ar thus it becomes arctan(𝑎𝑟 + 𝑎) − (𝑟 + 1)) − arctan ( 𝑛𝑟 𝑛+1 𝜋 𝜋 𝜋 2 4 4 )= − = Question #10 Prove that lim [ln ( 𝑥→0 1+cos 𝑥 1−cos 𝑥 )] ∼ 2 ln|𝑥| Answer: 1+cos 𝑥 log 1−cos 𝑥 (1+cos 𝑥)2 = log sin2 𝑥 = 2 log(1 + cos 𝑥) − 2 log|sin 𝑥| The first term is constant, while the sine is asymptotic to |𝑥|( for 𝑥 → 0) Question #11 How to sum up this inverse function lim ∑𝑛𝑟=0 tan−1 ( 𝑛→∞ 2𝑟 1+2𝑟 4 ) Answer: I'll leave it to you to prove by induction that the partial sum is arctan (1 − 1 𝑛2 +𝑛+1 ), so the limit is 𝜋 4 . One approach to obtaining this partial sum is that of hint ((𝑥, 𝑦) = (𝑓(𝑟 − 1), 𝑓(𝑟)) where 𝑓(𝑟) = 2𝑟 2 + 2𝑟 + 1) [arctan(2𝑟 2 + 2𝑟 + 1)]𝑛−1 = arctan 𝑛2 +𝑛 𝑛2 +𝑛+1 Edit: just to spell it out, the definition 𝑓(𝑟) ≔ arctan(2𝑟 2 + 2𝑟 + 1) gives 2𝑟 ∑𝑛𝑟=0 arctan ∑𝑛𝑟=0(𝑓(𝑟) − 𝑓(𝑟 − 1)) = 𝑓(𝑛) − 𝑓(−1) 4 = 1+2𝑟 where the second = can be proven by induction; this is the basis of any proof with telescoping series. In particular 𝜋 𝜋 𝜋 lim (𝑓(𝑛) − 𝑓(−1)) = arctan ∞ − arctan 1 = − = 2 𝑛→∞ Question #12 Evaluating lim 𝑥 2 −ln(cos 𝑥) 𝑥 𝑥→0 Answer: Without L'Hopital: let 𝑓(𝑥) ≔ 𝑥 2 − ln(cos 𝑥) . Then lim 𝑥 2 −ln(cos 𝑥) 𝑥→0 = lim 𝑥→0 𝑥 𝑓(𝑥)−𝑓(0) 𝑥−0 = = 𝑓 ′ (0) = 0 Question #13 Compute the limit: lim 1−cos3 (𝑥) 𝑥≥0 𝑥 sin(2𝑥) Answer: lim 1−cos3 (𝑥) 𝑥≥0 𝑥 sin(2𝑥) Apply L'Hopital's Rule lim 1−cos3 𝑥 𝑥→0 𝑥 sin(2𝑥) = lim 3 cos2 𝑥 sin 𝑥 𝑥→0 sin(2𝑥)+2𝑥 cos(2𝑥) 4 2 Again apply L'Hopital's Rule lim 𝑥→0 3(− sin(2𝑥) sin 𝑥+cos3 𝑥) 4 cos(2𝑥)−4𝑥 sin(2𝑥) = 3 4 Question #14 Evaluating trig limit lim √1−cos(𝑥 2 ) 𝑥→0 1−cos(𝑥) I have tried to simplify the expression using the identity 1 − cos(𝑥) = 2 sin2 (𝑥/2), but I have still failed to remove the indeterminate form. Answer: Hint: 𝛼 1 − cos 𝛼 = 2 sin2 2 Apply this with both numerator and denominator. Then use lim 𝑥→0 sin 𝑥 = lim 𝑥 𝑥 𝑥→0 sin 𝑥 =1 Question #15 tan(sin(𝑥)) Limit of sin(tan{(𝑥))} when x approaches 0 Answer: HINT By standard limits use that tan(sin 𝑥) sin(tan 𝑥) = tan(sin 𝑥) tan 𝑥 sin 𝑥 𝑥 sin 𝑥 sin(tan 𝑥) 𝑥 tan 𝑥 Question #16 How to calculate the limit lim sin −1 ( x)  x →1 The function takes the value 1 at x=1 but while approaching 1 from the left side, it takes the value 0 at all points. The function is not defined at x>1 so the right limit does not exist. The book mentions 1 as the answer as f(1)=1 but i don't understand why. Answer: 𝜋 Note that arcsin(1) = , not arcsin(1) = 1; this means in particular that for values x to the immediate left 2 of 1, also arcsin(𝑥) > 1 . Thus, the function floor(arcsin(𝑥)) is continuous at 1. However, the answer in the book is bad if it doesn't mention the fact that the function is continuous near x-value 1 and just gives the answer. Question #17 lim ℎ→0 (1+ℎ)12 −1 ℎ represents the derivative of the function f(x)=12 at the number a=12 Answer: Hope, it will help you Question #18 Can someone please explain to me why the following identity is true? 𝑎 𝑥 lim (1 + ) = 𝑒 𝑎 𝑥→∞ 𝑥 Answer: You have to recall the fundamental limit 1 𝑥 lim (1 + ) = 𝑒 𝑥→±∞ 𝑥 Think of it as a general rule like this: 1 ⋆ lim (1 + ) = 𝑒 ⋆→±∞ ⋆ where the star can be substituded by any expression (which tends to ±∞ ). So 𝑎 𝑥 𝑥⋅𝑥 𝑎 𝑥 1 𝑎 lim (1 + ) = lim [(1 + 𝑥 ) ] 𝑥 𝑥→±∞ 𝑥→±∞ = 𝑒𝑎 𝑎 Question #19 Can't figure out easy limit: lim+ sin(2𝑥 2 ) 𝑥→0 𝑥3 I was trying to help someone with their intro calc homework but got stumped by this super easy problem. I know we can apply L'Hôpital's rule to get lim+ 4 cos(2𝑥 2 ) 3𝑥 𝑥→0 which goes to 4 0 = + ∞ but I think this doesn't count as a proof. What am I forgetting? Answer: You did not forget nothing. (sin(2𝑥 2 ))′ = 4𝑥 cos(2𝑥 2 ) Note lim ( 𝑥→0 sin(2𝑥 2 ) 2𝑥 2 ⋅ 2𝑥 2 𝑥3 ) Thus, our limit it's +∞ for 𝑥 → 0+ and its −∞ for 𝑥 → 0− Question #20 Can't get out of a lim 0 0 indeterminate form for this limit (2𝑥 tan 𝑥) 𝑥→0 (1−𝑒 𝑥 )2 Answer: Set 𝑔(𝑥) = 2𝑥 tan(𝑥) ℎ(𝑥) = (1 − 𝑒 𝑥 )2 , 𝑓(𝑥) = 𝑔(𝑥) ℎ(𝑥) Then as lim 𝑔(𝑥) = lim ℎ(𝑥) = 0 you have that 𝑥→0 𝑥→0 𝑔(𝑥) 𝑔′ (𝑥) = lim ′ 𝑥→0 ℎ(𝑥) 𝑥→0 ℎ (𝑥) lim if the right hand side exists by l'Hôpital. But as lim 𝑔′ (𝑥) = lim ℎ′ (𝑥) = 0 you have that 𝑥→0 lim 𝑔′ (𝑥) 𝑥→0 ℎ′ (𝑥) = lim 𝑥→0 𝑔′′ (𝑥) 𝑥→0 ℎ′′ (𝑥) if the right hand side exists by l'Hôpital. But by evaluating this you can see that lim 𝑔′′ (𝑥) 𝑥→0 ℎ′′ (𝑥) 4 = =2 2 And therefore you have your limit.