How do I find this limit? I'm also pretty sure my 3rd step isn't correct. Reddit r/calculus

how do I find this limit I'm sure I have to  use lito rule but I don't know how to apply   it here I'm also pretty sure my third step is  incorrect and let me take a look at your third   step all right so let's have a look so here's the  limit and a and b are positive and first you put   this to be y okay and then I think because you  notice that we have natural log so you want it   to exponentiate both sides so e to this power  and E to that power right and you got that but   here's one small thing you're supposed to write  down the limit as X approaching Infinity so don't   drop that but this is right here is not the major  issue the next step is the major issue so right   here you had e to the Ln Ln and they just cancel  out the E and LNS this right here is actually not   correct no okay don't do that here is the D so let  me just make a note on the side when we have e to   the Natural laog if it's just one natural log of x  and assuming X is positive we're not talking about   complex numbers here assuming X is positive then  in this case yes you can cancel out the E and the   Ln and just get the X back this is okay but if  you have e to the Ln of X over Ln of Y in fact   there's no nice way to simplify this and that you  can use the change of Base formula and you write   Ln of X over Ln of Y as log base y of X but that  doesn't do us too too much good because the base   distance that will not be the same if you have  Ln Ln here and you say this is equal to X over   y because you cancel the E and Ln no this is not  correct so that is incorrect so the rest we don't   have to look at it and now the question is how do  we take of this limit well don't worry let me show   you just like you said we will be used Laing to  plus Rule and the reason why is because if we put   Infinity into this x b is positive so positive *  infinity is infinity plus one is still Infinity L   infinity is infinity so we have Infinity on the  top and then if you put Infinity on the bottom   we also get infinity similar reason so we have an  Infinity over Infinity case this is the indicator   that we get to use laal rule so this will be the  limit as X approaching Infinity to use Laos rule   we look at the top and take the derivative of it  and let's just focus on that first derivative of   Ln we just get one over the inside so 1 over BX  + one and don't forget the chain rule look at   the inside multiply by the derivative of BX + 1  the r of BX is just B the RO of plus one is just   zero so that's it and and we look at the bottom  take the derivative of it as well similar thing   right derivative Ln of whatever is just one  over that so ax^2 + 3 and then use the chain   rule derivative of ax s use the power rule put  the two to the front minus one so we get 2 a x   and the rity of plus three is just zero so this is  what we have and then we just have to simplify it   and and hope for the best so this right here is  going to be the limit as X approaching infinity   and now on the top we have B over b x + 1 and  then for the bottom here we can put this on   the top but remember we're dividing this so you  multiply by the reciprocal this goes to the top   so we will have times that which is ax² + 3 this  on the bottom so divided by 2 a x just like that   all right now just kind of multiply this and use  parentheses if necessary like this multiply this   out we get the limit as X approaching infinity and  distribute it we get B * that a b x² and then plus   3 B over distribute 2x so 2 a b x and x we get X  squ and this times that + 2 a x yeah now we have a   rational case when X approaches Infinity you have  two ways to do it one you can still use laitos   rule or two you can use the shortcut the shortcut  is just pay attention to the highest power on the   top and also the highest power on the bottom and  you compare them if your teacher let you to do it   then you do it if not I'll show you guys another  way too so in fact this right here we just have to   care about this and then care about that because  they both have X squ yeah so this doesn't matter   and again X is be it's because X is approach  Infinity so you get to do this now you see   this right here pretty much just gives you the  limit as X approaching Infinity ax a BX 2 over   2 a b x² here and the beauty of this is this and  that they all cancel it's just 1/ two no more X   so finally the answer is just one over two just  like that so if you in my class in fact I will   let I let my students do this so yeah that's it  but if your instructor does not let you do this   then you do the following way all right you can  still use ly to but just check this out another   way to do this is here we we have the limit as  X approaching Infinity I'm going to divide every   term by the highest power of X that we see which  is x s so here we have a b x² + 3B yeah I'm going   to divide this by x s I'm going to divide this by  X squ on the bottom we have 2 a b x² + 2 a x yeah   I will divide this by X squ I'll divide this by  x s just like that why you will see this and that   cancel and this and that cancel and we can reduce  that this becomes limit X approaching Infinity we   have AB plus 3 B over x² and then this right  here is 2 a b and then this right here is plus   this is not reduce yeah so 2 a but over X now  X goes to Infinity a stays by this term 3B over   infinity will go to zero this right here stays 2  a over infinity this right here will go to zero   so we just get AB over 2 a and we can reduce that  Final Answer 1 over two just like that hopefully   this right here is helpful and check out my  other videos for more D tutorials that's it

Channel: bprp calculus basics

Views: 19,983

Keywords: calculus limits, math tutorial, calculus tutorials, l'hopital's rule, bprp calculus

Id: i37Yr2PeX9M

Length: 7min 29sec (449 seconds)

Published: Wed Feb 07 2024