5

Determine the limits:lim T-5 (I -5)2 lim m cot(z) 040Vz? + 7 _ lim 1-3 I _ 3 2 COs 31; lim 170 e 12 _ 9 lim 1--3 2x2 + 7c+3 12 2x lim 1-2 - 12 _ 4r+4 (2+h)3 _ 8 lim...

Question

Determine the limits:lim T-5 (I -5)2 lim m cot(z) 040Vz? + 7 _ lim 1-3 I _ 3 2 COs 31; lim 170 e 12 _ 9 lim 1--3 2x2 + 7c+3 12 2x lim 1-2 - 12 _ 4r+4 (2+h)3 _ 8 lim h-0 sin 0 lim 0-0 0 + tan 0 lim In(z? 9) 1-3+lim Vr2 +3r I-022 + 21) HINT: conjugates

Determine the limits: lim T-5 (I -5)2 lim m cot(z) 040 Vz? + 7 _ lim 1-3 I _ 3 2 COs 31; lim 170 e 12 _ 9 lim 1--3 2x2 + 7c+3 12 2x lim 1-2 - 12 _ 4r+4 (2+h)3 _ 8 lim h-0 sin 0 lim 0-0 0 + tan 0 lim In(z? 9) 1-3+ lim Vr2 +3r I-0 22 + 21) HINT: conjugates



Answers

Use the basic limits $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ and $\lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0$ to find the following limits: (a) $\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}$ (b) $\lim _{t \rightarrow 0} \frac{\sin t}{4 t}$ (c) $\lim _{x \rightarrow 0} \frac{\cos x-1}{5 x}$ (d) $\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x^{2}}$

Use the basic Clement's limit as X approaches zero of sine X over X equals one under limit as X approaches zero close on the smartest one over X equals zero to find the following limits. Hey, B c d. All right, let's start with Hey, is the limit as t approaches zero of two t over scientific. Well, this looks kind of familiar, like this limit the first basic limit, except it's flipped around. Let me introduce you to the limit. Reciprocal you that isn't the limit. Reciprocate. So if you have the reciprocal of a limit that we know like T will do the next like X over sine X, then this is equal to the reciprocal of the limit. So we have sine x the limit of sine X over X equals one. What's the reciprocal of one? Well, which one? So the limit has ext approaches zero of the reciprocal function. The flip diversion of sine X of Rex is also one you can just flip with. All right, that's all fine and dandy. So now that we know that we can proceed with our problem So we know that the reciprocal this t over signed teething that is, one senses a constant, so we can just bring it out front. So now we have tea terms. A limit has t approach zero of tea over 70 and we know that this whole thing goes toe one. So we have two times one, it is equal to teach, So this whole limit at the beginning is equal to two. Perfect. Next step B B is the limit as X approaches zero co sign X squared minus one over Texas. So in this case, everything looks pretty normal. It looks exactly like this, except our variables are X squared and X squared instead of accidents. But that is fun because our variables are matching. You can just say that you'll find X squared minus one over X squared is equal to zero variables that you could make this happen. You could make it more formal, more math work By saying that X squared equals t you could make up your new variables. Let's say, yeah, you're making up new variables and then you would have co sign t over teeth, and that looks exactly like what we have. So that is definitely next question number. C letter C letter C is a limit. Experts zero of sign six X over sign five x Now there are several ways to find this and one of them low p Tall jewel is easier than doing this, but we were asked to use the basic Linda rules, so I'll show you a little trick. We've got signed six x and sign five X. That's like the top part of the sine X over X. We can't really do anything because he's got two sides. But what we can do is multiplied by a special form of one to make it look more like this. And that special form of one is something you might not have seen before one over X over one over X. So now we have something new that looks like this. Signed six x for X over sign five X x. Now you can probably handle them, right? Remember that you need the variables to be the same. And how do we do that while we can multiply the top and the bottom by special forms of one again? So this weaken do 6/6 to give us six Sign six X favor six x and we can multiply this one by 5/5 to give us five sign six five x over. Now there's a special rule, the limit as expertise, age of a function. He could do the limit of the top and the bottom one faction, this is the other. The women of the whole thing is equal to the limit of the top over the limit of the bus. So we can do that. We can take the limit. We can take a limit of the tough as X approaches zero and we could take the limit of the bottom as X approaches zero and then you can put the one over the other. And that's the end. So, my friends, what is a limit of the top? But we've got something that isa this sign six x over X and that goes toe one about this constant outfront six times one six. Then seeing on the bottom you've got this. Let me recognize Sign five experience by that that goes to one of this constant outfront, which gives us five. So the limit of the top is six over the limit of the bottom, which is five. So the answer is six defense. Last question is the limit. As X approaches zero of 10 to exit bricks, the limit is expected view of Shantou X. Now we know that both off the basic limits that we're using have a co sign and a sign in them No tan. So maybe we can try and put this 10 into something that we recognize a little bit better. Tan is the same thing assigned X over coastline next. But since Mrs Tan two X, it's like writing limit has X purchase zero of signed two x over co sign two X all over. Remember that you can keep change foot so we can re like this as a sign two x over x co sign two x meet This looks a little more manageable, but still I noticed that this is sort of close to the sine X over X limit the one that I want, but not quite. So I'm going to rewrite it again. Signed to its over X times one over show signed to This just looks better. The limit as approaches you. We still can't do this limit. We need this to be the same as that How do we do that multiplied by a special form of one? How about 2/2 when we did Ashley. What comes to Sign two x two X Times Forward Co sign. Remember that you can take the limit of one part from the one of the other and multiply them together to get the full limit of the entire function. So we know that this portion signed two x over two X goes to one from the limit X X approaches. Zero got to his constant, too. Times one and then we have the other part. One over coastline two X Here we can just plug in. Zero. What is the limit of one over coastline? Two. X as X approaches zero. Well, it's like saying one over coastline of zero coastline of zero is one perplexing 1/1 for this whole thing is just times a big one. So basically out of the picture, looking back at this part but are two times one times one thing is the one thing over here. Two times two times 11 from over here, because your whole limit

Okay, so we see that we have to Paula. No meals here. And this is a rational expression. So it's put up or limit into two limits. One off our numerator and another of our denominator. Okay, I know. We see that we have to polynomial is so it gives just plug in those polling. No meals at our points. Accessible to to that is we get to one. It's five over two squared, plus four. So that's equal to negative. 3/2 squared plus four, which is Bates. So we get negative 3/8 as our solution.

Okay, So if you used jets up to solve Fergus Limit, we see that we have rigid, right girl. So it's charged to, um, change this formula. So why don't we try? And that tangent of two X will be right that in terms of signing coastline so that actually that sign of two x over co sign of two X and multiplied after I won over five X will recall that if we have sign of X over X and we take the limit as expert usual, that's equal to one. So let's rewrite Well, you have will write it in terms of sign over its. So we have sign of two acts over five x and end times when over coastline of two X. Do you see that this limit here, that's one. So I went over five times one times one over co sign of zero, and that's also equal to one. So we shed our limits. That's equal to 1/5

Brooklyn. Number 40. The limits off extends to zero. Oh, then that's the access. Where? Well, sign square. Dr X Over. It's just weird. Ah, which can beat it. And as then three x were over. Its is where us? So it we're five. It's over. Excess were which can be also written as three times sign. It's where the XT Where over three Xs. Where times one over or sign. But he excess were plus 25 Sign where or five x over 25. It's extra square. Ah, the limit extends to therefore signed Square three. Answer. Spread over three. X squared is equal to one, and the limits were went over. Society X squared when extends to zero, is equal to one plus 25 times the limit for science. Weird five x over. Ah 25. It's where 25 in its worst and beating. And that's fine. X Would where so science were five X over five x. Older square is equal to one square, so the thunder answer would get three times 25 6 20 Hey,


Similar Solved Questions

5 answers
HIC.25.# The equation of motion of a rocket-- propelled sled is, from Newton'$ law; mi = f CU where m is the sled mass, fis the rocket thrust, and c is an air resistance coefficient: Suppose that m 1000 kg and c 500 N that v(0) slm: Suppose also 0 and f = 75,000 N for t = 0. Determine the speed of the sled at t = 10 s.
HIC. 25.# The equation of motion of a rocket-- propelled sled is, from Newton'$ law; mi = f CU where m is the sled mass, fis the rocket thrust, and c is an air resistance coefficient: Suppose that m 1000 kg and c 500 N that v(0) slm: Suppose also 0 and f = 75,000 N for t = 0. Determine the spee...
5 answers
A proposed mechanism for the reaction 2NO(g) +2H2(g) _ N2(g) + 2HzO(g): Step 1: 2NO(g) _ N202(g) (very fast, reversible) Step 2: N2Oz(g) + H2(g) _ NzO(g) H2O(g) (slow) Step 3: NzO(g) Hz(g) ~Nz(g) + H2O(g) (fast) What is the rate law for the overall reaction?K[NO]1/2[ [Hz] k[N2O][Hz] k[NO]2 K[NO]?[Hz] k[NO]2[Hz]2
A proposed mechanism for the reaction 2NO(g) +2H2(g) _ N2(g) + 2HzO(g): Step 1: 2NO(g) _ N202(g) (very fast, reversible) Step 2: N2Oz(g) + H2(g) _ NzO(g) H2O(g) (slow) Step 3: NzO(g) Hz(g) ~Nz(g) + H2O(g) (fast) What is the rate law for the overall reaction? K[NO]1/2[ [Hz] k[N2O][Hz] k[NO]2 K[NO]?[H...
5 answers
(20 points) Use the Root test to determine whether the series converges or diverges~Zn32nHint: In the Lecture note 4, see Example 6 on Slide 16, and Exercise 27 on Slide 18.
(20 points) Use the Root test to determine whether the series converges or diverges ~Zn 32n Hint: In the Lecture note 4, see Example 6 on Slide 16, and Exercise 27 on Slide 18....
5 answers
Find dBz(0, 0, 0) , the z component of the magnetic field at the point € = y = 2 == 0 from the current [ flowing over a short distance dl dl j located at the point Tc == %1 i . (Figure 3) Express your answer in terms of [, Tl, [lO, T,and dl. Recall that a component is a scalar; do not enter any unit vectorsView Available Hint(s)AzddBz(0, 0,0)
Find dBz(0, 0, 0) , the z component of the magnetic field at the point € = y = 2 == 0 from the current [ flowing over a short distance dl dl j located at the point Tc == %1 i . (Figure 3) Express your answer in terms of [, Tl, [lO, T,and dl. Recall that a component is a scalar; do not enter an...
5 answers
CiUsersyhp Desklop/Assigrment9201-I6SMAT3119520-Fall%20209620K1) pdfRead &loudLaeHigtlightWucellulMarks)Kicnin AlcenngomneelnunSmIt high I6" cote mn hus jet-pow ered ske; The CDfI (orce of 280 purllcl Ine Wuunce ol ine Eope cunhirat ofskis and Kicrn 50 kg Uha Ivel = negligblel Kieran'> spced Mt tx bullomn Aum > What i the FckalGnkat- Inctlon ohi SL UDI iot (Draw Inre tody dugrimt)
CiUsersyhp Desklop/Assigrment9201-I6SMAT3119520-Fall%20209620K1) pdf Read &loud Lae Higtlight Wucellul Marks) Kicnin AlcenngomneelnunSmIt high I6" cote mn hus jet-pow ered ske; The CDfI (orce of 280 purllcl Ine Wuunce ol ine Eope cunhirat ofskis and Kicrn 50 kg Uha Ivel = negligblel Kieran&...
5 answers
Skill 57: Gas Law Applications Acetylene burns according to the following reaction: 2 CzHz (g) 5 Oz (g) 4 CO2 (g) + 2 HzO (g)Change in ng Product ng Reactant ngdecrease or remain nearly constant as the reaction Does the volume increase or proceeds? Explain.
Skill 57: Gas Law Applications Acetylene burns according to the following reaction: 2 CzHz (g) 5 Oz (g) 4 CO2 (g) + 2 HzO (g) Change in ng Product ng Reactant ng decrease or remain nearly constant as the reaction Does the volume increase or proceeds? Explain....
5 answers
Question 161 ptsfront ticr seat and $70 tor back tier seat: Ticket sales Ticket prices for concert were 5100 for wcre 800 front tier scats and 1,200 back tier scats The weighted mean price was87.5082.0092.50none of the other choices are correct85.00
Question 16 1 pts front ticr seat and $70 tor back tier seat: Ticket sales Ticket prices for concert were 5100 for wcre 800 front tier scats and 1,200 back tier scats The weighted mean price was 87.50 82.00 92.50 none of the other choices are correct 85.00...
5 answers
Provide proper IUPAC names for each compound below. 22. (CHJCHCH2CHZCH(CH2CH3)CH2C(CH3)323.CH; H;CCHCH-CCH CHCH,CH; CHs C(CHsh2 -
Provide proper IUPAC names for each compound below. 22. (CHJCHCH2CHZCH(CH2CH3)CH2C(CH3)3 23. CH; H;CCHCH-CCH CHCH,CH; CHs C(CHsh 2 -...
5 answers
Question 21 (2 points) What is the IUPAC name of following compound?
Question 21 (2 points) What is the IUPAC name of following compound?...
5 answers
A is experiencing a deceleration directly proportional to its speed, while it is moving toward the right: &4 = -0.SvA (VA is the instantaneous velocity of A)The initial position XAO =0 and the initial velocity of A VA =[Om/s when t =0 determine (a) A's velocity when XA becomes 2 m and (b) B*s velocity when A is located at XA 2 m and (c) B's velocity when t = 5 seconds_
A is experiencing a deceleration directly proportional to its speed, while it is moving toward the right: &4 = -0.SvA (VA is the instantaneous velocity of A) The initial position XAO =0 and the initial velocity of A VA =[Om/s when t =0 determine (a) A's velocity when XA becomes 2 m and (b)...
5 answers
Supeota U= Jestoon fynnghtouchthe # eornttkts Vorceton a)end tleoly vt69r 02 4 0 Anhlrata | tha latcd Untnn ' Mlkanla Kuleeeln {ostu celt (eltdBahdoninm;
Supeota U= Jestoon fynnghtouchthe # eornttkts Vorceton a)end tleoly vt69r 02 4 0 Anhlrata | tha latcd Untnn ' Mlkanla Kuleeeln {ostu celt (eltd Bahdoninm;...
5 answers
5) (12) Let Tcz, T={0,3,8,15,24,35} Prove that set T is countable O1 uncountable by finding a one to one correspondence
5) (12) Let Tcz, T={0,3,8,15,24,35, } Prove that set T is countable O1 uncountable by finding a one to one correspondence...
5 answers
The formation constant* of [M(CN) 6 ]4− is 2.50×1017 , where Mis a generic metal. A 0.140 mole quantity of M(NO3)2 is added to aliter of 1.040 M NaCN solution. What is the concentration of M2+ions at equilibrium?
The formation constant* of [M(CN) 6 ]4− is 2.50×1017 , where M is a generic metal. A 0.140 mole quantity of M(NO3)2 is added to a liter of 1.040 M NaCN solution. What is the concentration of M2+ ions at equilibrium?...
5 answers
Suppose that we have a population with proportion P = 0.62 and arandom sample of size n = 900 drawnfrom the population. Keeping you answers to 3 decimal places,finda. the probability that the sample proportion is between 0.60and 0.64?b. the probability that the sample proportion is between 0.57 and0.67?
Suppose that we have a population with proportion P = 0.62 and a random sample of size n = 900 drawn from the population. Keeping you answers to 3 decimal places, find a. the probability that the sample proportion is between 0.60 and 0.64? b. the probability that the sample proportion is between 0.5...
5 answers
15. Given 3.01 of an ideal gas in a 73.6 dm3 container at 25C Calculate the internal energy change for a reversible pressure increases to 2.5 atm at constant initial volume given Cp,m = 49.884 J/mol.
15. Given 3.01 of an ideal gas in a 73.6 dm3 container at 25C Calculate the internal energy change for a reversible pressure increases to 2.5 atm at constant initial volume given Cp,m = 49.884 J/mol....

-- 0.020969--