Converging and Diverging Sequences Using Limits - Practice Problems

[@TheOrganicChemistryTutor]

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Nursultan Beloved this is only the first section in the first chapter in my course and at most only one question will come on it in the test.

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@@theadel8591 I feel like that'll be the case for us, except I bet most of it is going to be power series

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Professor Organic Chemistry Tutor, thank you for using multiple well-known examples to explain/analyze Converging and Diverging Sequences in Calculus Two. In all Calculus textbooks, there is a whole chapter on Sequences and Series. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

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[@horrornight8240]

Yes it is cause the lim as n>infinity of ln(1+(1/n))/(1/n) is ln(1+0)/(0) which is undefined.

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[@RA-pu9jo]

for anyone else scratching there head at this! plugging in inf into (1+1/n)^n = (1+1/inf)^inf = (1+0)^inf = (1)^inf = 1^inf = undefined this is why we had to solve it

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[@mynameisgrand8937]

At 4:30 why do you multiply the top and bottom by 1/n? How do I recognize when to use this?

[@FerghusCameron]

Did you find out the reason? I just started Pre calculus and would like to know thr proof for multiplying the top and bottom by 1/2

[@hungry3166]

​@@FerghusCameron Just to take the n to an easier place. This can be solved by many ways, and you can keep the n and evaluate it to find the answer (if it's converging or diverging) but dividing by 1/n takes the n to 5/n which leads to the final answer easily (because of the rule 1/n=0).

[@joshuafleckenstein351]

@@FerghusCameronif you took the limit as n approaches infinity, you’d end up with infinity over infinity. By multiplying by the numerator and denominator by 1/n, you are able solve. Another method of solving for the infinity/infinity problem is applying L’Hopital’s Rule, which would also give you the same answer and is more applicable in vague situations where multiplying by 1/n isn’t available or intuitive.

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[@cyberzod99]

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[@benhigh1910]

why would you add the last step at 22:15? 1/(n+2) is just 1/infinity which is zero. Absolutely no need to multiply top and bottom to get the zero in the numerator

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[@ermiastadesse7042]

why do u use l'hopital rule insted of just substitute infinity in the equation lim n-> infinity n * sin(1/n)??? and all your video are just unreal. u made life easy for us.

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[@spammusubi1607]

Wait so at 22:55 why do you multiply by 1/n instead of just evaluating the limit there??

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[@vestia8444]

for the question at 13:44 couldn't you just simplify sin(n)/n to 1? Is that an identity that can be used?

[@nabilah2510]

for 13:21 why can’t we use lopital rule and make it lim cos(n) = 1?

[@hrushikeshnaik2968]

At 5:10 why was the 1/n adjustment done?

[@jahmallidge5115]

It was just to avoid L'hopital's rule. It doesn't change the limit if you multiply the top and bottom of the fraction by any real number other than zero, infinity and negative infinity.

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[@Blockwise2]

Should not we apply stolz-cesara instead of l’hospital when dealing with sequences? Please explain!

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[@masterofdisaster1911]

7:10 even though it seems counter-intuitive it is technically a continuous function given the ε-δ definition of continuous functions.

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[@isaackim9181]

Could you not have used L'Hopital's rule for the limit as n approaches infinity of a_n = 8n/3n-5? (5:35)

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What should you do if it goes to 0 with infinity but its limit for negative infinity is infinite? Does it till converge?

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[@mansafabrar6872]

Will give u a good idea about convergence nd divergence. But formaly the idea of Convergence nd divergence is different.

[@heli7147]

At 13:49, why can you determine it's convergent since lim is 0 here? The divergence test on our slide said that you can't determine whether it's convergent or divergent straight forward if lim=0. I'm so confused:(((

[@footballmad2790]

I think you're getting confused between the convergence and divergence of series and of sequences, the divergence test of series needs the limit you're talking about, and in this example we can't determine whether the infinite sum is convergent or divergent(the series) but what we can determine in this example is that the sequence will end up at 0 when we approach infinity

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[@EzriMata]

I'm lost at 19:08. Apparently the answer is now 1, but why couldn't it be 0? sin(1/n) would return 0 and anything multiplied by 0 is 0. Why is that invalid? Are we only able to find the sequence when n only appears once in the function?

[@EzriMata]

Never mind i understand now. The sin of 1/n may approach zero but it never reaches zero. Then it is multiplied by what ever n is as it approaches infinity.

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