Converging and Diverging Sequences Using Limits - Practice Problems

[@TheOrganicChemistryTutor]

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[@georgesadler7830]

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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[@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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[@mansafabrar6872]

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

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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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[@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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[@Blockwise2]

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

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[@imbored5951]

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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[@spammusubi1607]

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

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For people who didn't get horizontal asymptote, find x in terms of y..

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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?

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