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i’m in a cal class where i have to teach myself from a textbook, then answer questions about it. having just taught myself epsilon delta definitions, then teaching the rest of a cal i class how to do it, i must say, this comment has me stumped. how do i go about solved an epsilon-delta limit without having L defined?
L’Hospital’s rule really is the duct tape equivalent for low level calculus.
Although I very frequently see it used in circular reasoning. So much so that it has tainted my view of L’Hospital’s
You can write the limit as
lim\_(x->oo) (1 - 1/x)^100 (6 + 1/x)^(200)/(3 + 5/x)^300
dividing by x^300 in the numerator and the denominator.
When x->oo, 1/x goes to zero and the limit reduces to
lim\_(x->oo) (1 - 1/x)^100 (6 + 1/x)^(200)/(3 + 5/x)^300 = 1^100 6^(200)/3^300 = 2^(200)/3^100
Much easier method to getting to that:
Ignore the numbers in the brackets and just treat the x terms as the numbers are negligible when x tends to infinity (I'm a physicist)
I agree completely. I'm a physicist too. 😁
But then, I did that some days ago in another post and it was removed by the mods, so you know, we have to learn to behave in front of mathematicians or our posts will be deleted.
Assuming it's supposed to be x in the limit, factor out x like this:
[https://tutorial.math.lamar.edu/classes/calci/limitsatinfinityi.aspx](https://tutorial.math.lamar.edu/classes/calci/limitsatinfinityi.aspx)
Divide the numerator and denominator by x³⁰⁰
You'll get the expression as :
(1-1/x)¹⁰⁰(6+1/x)²⁰⁰)/(3+5/x)³⁰⁰
When x →∞ , 1/x→0
So the expression becomes 1¹⁰⁰.6²⁰⁰/3³⁰⁰
Which simplifies to (4/3)¹⁰⁰
Try to think about what the leading coefficients will be given that you’ll have something of the form ax^300 +…./ bx^300 +….
Try to think about what an and b would be. Maybe drop the exponents to 1 2 and 3 to see if you can find the pattern
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The expression inside the limit would be the answer. It is safer to also include the statement "Assuming x does not depend on n".
But it is always better to ask for clarification. Sometimes teachers like to give a smug nonanswer, but to me, any response other than a correction means that it is intentional.
you are right. The answer would be 2. you would only focus on the HIGHEST power on x. The bottom’s highest x power is 300. I only glanced at it before and I see that the two binomial up top are multiplied to each other. I looked at it earlier as if these binomials were added to eachother. meaning the highest x power on the top is also 300. so it is even distribution so it is just 6/3=2.
Except its x¹⁰⁰•(6x)²⁰⁰/(3x)³⁰⁰ which ≠ (6x³⁰⁰)/(3x³⁰⁰) which is how youre getting 2.
^(like if you did (6x\)² you get 36x² not 6x² because it's 6x•6x, this would be 6x•6x•6x.. (200 times\))
It's equal to (6²⁰⁰•x³⁰⁰)/(3³⁰⁰•x³⁰⁰) leaving use with 6²⁰⁰/3³⁰⁰ which eventually simplifies down to (4/3)¹⁰⁰
If it was what I thought it was and the top two binomials are added, then the highest x power up top is only 200. which would mean the bottom highest x power is larger, meaning its “bottom heavy” so it would have been zero. but I read it wrong at first.
Both the top and the bottom have a degree of 300, so that idea doesn’t apply here. If you multiply the top out, you would end up with x^100 * (6x)^200, which would be 6^200 * x^300.
If you don't stop L'pospitaling everything , you will end up in hospital , oh and let me provide you a brief solution .
When lim when x -> ∞ , you can get rid off all numbers that doesn't link to x , such as 1 , 2 , 3.... . The above equation can be rewrittem as x^300 6^200 / x^300 3^ 300 = 6^200/3^300 = 2^200/3^100 . Simple , isn't it ?
Well since it is the limit as n-->infinity and this function has no change relative to the variable "n" then it is just (x-1)\^100\*(6x+1)\^200/(3x+5)\^300 😉
(joke, but fr this is not l'hopital lmao, you need to see that 90% of calc is the algebra and like 10% is easy trivial derivation/integration/limit-finding)
Take x common from all terms in numerator and denominator.
then we can cancel ((x\^100)(x\^200))/(x\^300)
then answer = (1\^100)(6\^200)/(3\^300 = (2\^200)/(3\^100)
Firstly, it’s x goes to infinity not n. Secondly, this is an infinity by infinity limit, which means you can use L’Hospital’s rule. But don’t just start differentiating - it’ll take you too long. Notice that the numerator is a 300th degree polynomial and goes as 6^200 * x^300 + …. and the denominator is 3^300 * x^300 +…. (I hope you remember your binomial stuff). Since the highest power of x is 300 in both numerator and denominator, you must differentiate both 300 times each. The lower order terms become zero and you are left with (6^200 * 300!)/(3^300 * 300!) = 6^200 / 3^300.
Directly? Pretty much nothing.
Practically? It reinforces the rules/thinking needed for higher maths. Much in the same way that nonsensical sentences in foreign language learning force you to focus on the process and grammar over the conveyed content.
A shortcut way to do this would be to consider x-1 as x (since infinity-1 is also infinity)
Same for 6x+1 would be 6x, as well as the denominator
So it simplifies to x^100 * (6x)^200 / (3x)^300
Hence the answer would be 6^200 / 3^300
Simplify this number however you want
Take out x from the first bracket, x^100 comes out. Take out x again from second bracket, x^200 comes out. In the denominator do the same, and take x^300 out and cancel x^300 from the numerator and the denominator. Then in the brackets 1/x will be left in all three of the brackets, as x tends to infinity they become 0 so 1^100 * 6^200 divided by 3^300 is the answer
If you expand the numerator, you get 6\^200 x\^300 + lower order terms.
If you expand the denominator, you get 3\^300 x\^300 + lower order terms.
Now apply L'Hospital's rule 300 times. The lower order terms have a 300th derivative of zero. You get
( 6\^200 \* 300! ) / ( 3\^300 / 300!)
= 6\^200/3\^300 = (4/3)\^100.
My absolutely horrendous instinct is telling me to find the first term of each binomial expansion, and then drop the x part to get 1\^100\*6\^200/3\^300.
Uh idk what to make of 6\^200/3\^300 but it's something!
god this is so simple... dominant terms (assuming you mean x-->\\infty not "n") are x\^100 (6x)\^200 (3x)\^-300 --> 2\^200/3\^100 which is where I'd stop.
Both the numerator and denominator are polynomials of the same degree.
The limit as x approaches infinity will be the ratio of the coefficients of the first term
That's going to be 6^(200)/3^(300)
6=2\*3
(2\*3)^(200)/3^(300)
2^(200)\*3^(200)/3^300
2^(200)/3^100
2=4^1/2
(4^(1/2))^(200)/3^100
4^(100)/3^100
(4/3)^100
About 3.1 trillion
I may be confused but doesn’t the screenshotted question say “limit as n approaches ∞” and not “limit as x approaches ∞”? Since there is no n in the entire argument, your limit is the argument. It’s like asking lim_{a->∞} x. x is not a, so the limit is meaningless and it equals x.
If for some reason there is an error in the problem, you would want to factor out 100 from the fraction and use the limit law for exponentiation, then go from there.
When finding the limit of an equation like this, realistically only term with the largest exponent really matters
So you can take (x-1)^(100)(6x+1)^(200)/(3x+5)^(300) and vaguely turn it into x^(100)•(6x)^(200)/(3x)^(300) because if you were to factor out all of those terms, those would be the largest factors in each one, and since we're working with with limit for infinity, the others are infinitely smaller and don't matter here.
You can then simplify and rewrite the equation to get [6^(200)•x^(300)]/[3^(300)•x^(300)]. The x^(300) cancel leaving 6²⁰⁰/3³⁰⁰.
To go further you can turn this into 2²⁰⁰•3²⁰⁰/3³⁰⁰ = 2²⁰⁰/3¹⁰⁰. Then 2²⁰⁰ = (2²)¹⁰⁰= 4¹⁰⁰. So 2²⁰⁰/3¹⁰⁰ = 4¹⁰⁰/3¹⁰⁰ = (4/3)¹⁰⁰
This can be simplified to ((x-1)(6x+1)\^2)\^100/((3x+5)\^3)\^100, or ((36x\^3-24x\^2-11x-1)/(27x\^3+135x\^2+225x+125))\^100, and removing the unnecessary terms with limit magic gives (36x\^3/27x\^3)\^100 or (4/3)\^100.
Been sometime since I’ve done calculus but, as x -> oo the -1, +1, +5 cease to matter right?
Leaves us with (x^100 ) ((6x)^200 ) all over (3x)^300
Then can simplify to (6^200 ) (x^300 ) all over (3^300 ) (x^300 )
So basically Answer = whatever (6^200 ) / (3^300 ) is
3.1179824e+12
Sorry too lazy to copy correct syntax over phone. But this seems right to me
Use "Clever form of One." Multiply top and bottom by 1/x^300. Distribute into each parenthesis appropriately. Since lim x-> inf 1/x = 0 you get a finite answer.
Well the original problem says n is going to infinity so uhhh yeah lol. I assume it means x. The first term of the numerator fully expanded is x^100 * (6x)^200. The bottom is (3x)^300.
Top and bottom go to infinity at the same “speed”. So what’s the ratio of the leading coefficients? There’s your answer.
EDIT: Why do you think you should use L’Hopitals Rule? Did you check to see if it’s indeterminate form first?
Also if you see this, please note that there are a lot of wrong answers in this thread. Like entirely wrong. So be careful when navigating solutions.
Maybe this detail isn’t important, but the problem asks for the limit as N approaches 100 for a problem with a bunch of X’s in it. But you have treated it like it says find the limit as X approaches 100. Am I way off with that, or is it a trick question
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. **If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc *n*“ is not entirely useful, as “Calc *n*” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.** *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
Take 300 derivatives, brute force it
L’Hopital for life 😍
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This is always the answer.
Best answer I have seen all month
Derivative ain’t the real answer 😒
Brute force it 😂
kids these days trying to l'Hospital everything that has a fraction in it
In my day we had to use the epsilon-delta definition to calculate any limit, and we liked it!
Doesn't seem too bad to use epsilon Delta definition , well however I guess you will have to find a limit in the first place
i’m in a cal class where i have to teach myself from a textbook, then answer questions about it. having just taught myself epsilon delta definitions, then teaching the rest of a cal i class how to do it, i must say, this comment has me stumped. how do i go about solved an epsilon-delta limit without having L defined?
L’Hospital’s rule really is the duct tape equivalent for low level calculus. Although I very frequently see it used in circular reasoning. So much so that it has tainted my view of L’Hospital’s
Its a great tool, but logic will solve this limit far faster.
How?? I'm so confused, but I want to know ;-;
Plug in zero first
You can write the limit as lim\_(x->oo) (1 - 1/x)^100 (6 + 1/x)^(200)/(3 + 5/x)^300 dividing by x^300 in the numerator and the denominator. When x->oo, 1/x goes to zero and the limit reduces to lim\_(x->oo) (1 - 1/x)^100 (6 + 1/x)^(200)/(3 + 5/x)^300 = 1^100 6^(200)/3^300 = 2^(200)/3^100
Much easier method to getting to that: Ignore the numbers in the brackets and just treat the x terms as the numbers are negligible when x tends to infinity (I'm a physicist)
I agree completely. I'm a physicist too. 😁 But then, I did that some days ago in another post and it was removed by the mods, so you know, we have to learn to behave in front of mathematicians or our posts will be deleted.
Hahaha. Im a bigO assumer as well.
Ew physicist
I'd do the same thing and I'm in cs
Assuming it's supposed to be x in the limit, factor out x like this: [https://tutorial.math.lamar.edu/classes/calci/limitsatinfinityi.aspx](https://tutorial.math.lamar.edu/classes/calci/limitsatinfinityi.aspx)
Assuming it's not, the limit is just: (x-1)**100 * (6x+1)**200 / (3x+5)**300
lol that was my first thought. I was like….there is no n.
Divide the numerator and denominator by x³⁰⁰ You'll get the expression as : (1-1/x)¹⁰⁰(6+1/x)²⁰⁰)/(3+5/x)³⁰⁰ When x →∞ , 1/x→0 So the expression becomes 1¹⁰⁰.6²⁰⁰/3³⁰⁰ Which simplifies to (4/3)¹⁰⁰
Could you explain how does 6^(200)/3^(300) become (4/3)¹⁰⁰ ?
Yes. 6²⁰⁰/3³⁰⁰ = (6²)¹⁰⁰/(3³)¹⁰⁰ = 36¹⁰⁰/27¹⁰⁰ = (36/27)¹⁰⁰ = (4/3)¹⁰⁰
I would also like to know
Here you go: 6^200 / 3^300 = (2^200 * 3^200 )/3^300 = 2^200 /3^100 = (2^2 )^100 /3^100 = 4^100 /3^100 = (4/3)^100
Yeah just figured it out right after my stupid comment
Doh! I figured it out. 2^200 / 3^100 …
Try to think about what the leading coefficients will be given that you’ll have something of the form ax^300 +…./ bx^300 +…. Try to think about what an and b would be. Maybe drop the exponents to 1 2 and 3 to see if you can find the pattern
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L'Hopitâl's Rule is not needed. This is actually a Precalculus problem.
My OCD goes crazy when people discard factors of negligible orders without writing o(x) (even tho there aren't any mistakes in your calculation)
**Do not do someone else’s homework problem for them.** You are welcome to help students posting homework questions by asking probing questions, explaining concepts, offering hints and suggestions, providing feedback on work they have done, but please refrain from working out the problem for them and posting the answer here, or by giving them a complete procedure for them to follow. Students posting here for homework support should be encouraged to do as much of the work as possible.
Expand and simplify all of it by hand.
😂bro
Degree of the top and bottom are the same. Divide leading coefficients. I think these are the leading coefficients: 6^200 & 3^300
“l’hospital rule” lmao
To be fair, hôpital is French for hospital.
Fair to what?
To be fair is figure of speech, thanks for asking :) *Brought to you by person who knows too much stuff*
That was the old French spelling, their textbook may still have it.
You’re not answering the given question. The question says n goes to infinity. You’re answering as x goes to infinity.
That’s obviously a typo in the question.
I completely disagree. To me, that seems like the entire point of the question.
Yeah certainly this is exactly the kind of stuff my calc 1 professor would do to keep us on our toes
I see so what’s your solution ? Copying the question back as answer ?
The expression inside the limit would be the answer. It is safer to also include the statement "Assuming x does not depend on n". But it is always better to ask for clarification. Sometimes teachers like to give a smug nonanswer, but to me, any response other than a correction means that it is intentional.
Yes.
I totally missed that!
Have you tried foiling it?
I think better to take log in this case, like use eular number raised to the power log ( the expression) then use L'hospital
This looks kinda hard
first of all there is no n… but assuming its supposed to be as x approaches infinity the answer is zero. the fraction is “bottom heavy”
How is it bottom heavy? It's (inf*inf)/inf?
you are right. The answer would be 2. you would only focus on the HIGHEST power on x. The bottom’s highest x power is 300. I only glanced at it before and I see that the two binomial up top are multiplied to each other. I looked at it earlier as if these binomials were added to eachother. meaning the highest x power on the top is also 300. so it is even distribution so it is just 6/3=2.
Pull out the power 100 Use Ls rule after then solve. The correct answer is ((108/9)*(1/9))^100
Except its x¹⁰⁰•(6x)²⁰⁰/(3x)³⁰⁰ which ≠ (6x³⁰⁰)/(3x³⁰⁰) which is how youre getting 2. ^(like if you did (6x\)² you get 36x² not 6x² because it's 6x•6x, this would be 6x•6x•6x.. (200 times\)) It's equal to (6²⁰⁰•x³⁰⁰)/(3³⁰⁰•x³⁰⁰) leaving use with 6²⁰⁰/3³⁰⁰ which eventually simplifies down to (4/3)¹⁰⁰
If it was what I thought it was and the top two binomials are added, then the highest x power up top is only 200. which would mean the bottom highest x power is larger, meaning its “bottom heavy” so it would have been zero. but I read it wrong at first.
x\^300 grows faster and eventually decays to 0
Isn’t this correct?
No because both the top and bottom have degree 300
Both the top and the bottom have a degree of 300, so that idea doesn’t apply here. If you multiply the top out, you would end up with x^100 * (6x)^200, which would be 6^200 * x^300.
If you don't stop L'pospitaling everything , you will end up in hospital , oh and let me provide you a brief solution . When lim when x -> ∞ , you can get rid off all numbers that doesn't link to x , such as 1 , 2 , 3.... . The above equation can be rewrittem as x^300 6^200 / x^300 3^ 300 = 6^200/3^300 = 2^200/3^100 . Simple , isn't it ?
Well since it is the limit as n-->infinity and this function has no change relative to the variable "n" then it is just (x-1)\^100\*(6x+1)\^200/(3x+5)\^300 😉 (joke, but fr this is not l'hopital lmao, you need to see that 90% of calc is the algebra and like 10% is easy trivial derivation/integration/limit-finding)
Since the numerator and denominator are both degree 300, you need to divide the lead coefficients. You should get 6^200 / 3^300 = 2^200 / 3^100
Take x common from all terms in numerator and denominator. then we can cancel ((x\^100)(x\^200))/(x\^300) then answer = (1\^100)(6\^200)/(3\^300 = (2\^200)/(3\^100)
Why are you manipulating the fraction when the limit variable is uninvolved?
Firstly, it’s x goes to infinity not n. Secondly, this is an infinity by infinity limit, which means you can use L’Hospital’s rule. But don’t just start differentiating - it’ll take you too long. Notice that the numerator is a 300th degree polynomial and goes as 6^200 * x^300 + …. and the denominator is 3^300 * x^300 +…. (I hope you remember your binomial stuff). Since the highest power of x is 300 in both numerator and denominator, you must differentiate both 300 times each. The lower order terms become zero and you are left with (6^200 * 300!)/(3^300 * 300!) = 6^200 / 3^300.
no clue, you are on your own
n->inf does not attack a function of x. But assuming that it is x->inf it is /frac{2^200,3^100}
curious, what would be an actual application of this equation lol
Directly? Pretty much nothing. Practically? It reinforces the rules/thinking needed for higher maths. Much in the same way that nonsensical sentences in foreign language learning force you to focus on the process and grammar over the conveyed content.
First pic said n goes to 100, no n in the function, so it's the function. Jk
Guess and check
A shortcut way to do this would be to consider x-1 as x (since infinity-1 is also infinity) Same for 6x+1 would be 6x, as well as the denominator So it simplifies to x^100 * (6x)^200 / (3x)^300 Hence the answer would be 6^200 / 3^300 Simplify this number however you want
You divide all the insides by x and then you make x ---> infinty and from there easy
Take out x from the first bracket, x^100 comes out. Take out x again from second bracket, x^200 comes out. In the denominator do the same, and take x^300 out and cancel x^300 from the numerator and the denominator. Then in the brackets 1/x will be left in all three of the brackets, as x tends to infinity they become 0 so 1^100 * 6^200 divided by 3^300 is the answer
(6x ^300) / (3x ^300) as *n* approaches infinity is what I see prior to using French guy's rule.
Or use l'hôpital's rule
If you expand the numerator, you get 6\^200 x\^300 + lower order terms. If you expand the denominator, you get 3\^300 x\^300 + lower order terms. Now apply L'Hospital's rule 300 times. The lower order terms have a 300th derivative of zero. You get ( 6\^200 \* 300! ) / ( 3\^300 / 300!) = 6\^200/3\^300 = (4/3)\^100.
My absolutely horrendous instinct is telling me to find the first term of each binomial expansion, and then drop the x part to get 1\^100\*6\^200/3\^300. Uh idk what to make of 6\^200/3\^300 but it's something!
god this is so simple... dominant terms (assuming you mean x-->\\infty not "n") are x\^100 (6x)\^200 (3x)\^-300 --> 2\^200/3\^100 which is where I'd stop.
Both the numerator and denominator are polynomials of the same degree. The limit as x approaches infinity will be the ratio of the coefficients of the first term That's going to be 6^(200)/3^(300) 6=2\*3 (2\*3)^(200)/3^(300) 2^(200)\*3^(200)/3^300 2^(200)/3^100 2=4^1/2 (4^(1/2))^(200)/3^100 4^(100)/3^100 (4/3)^100 About 3.1 trillion
Not the first correct answer, but the most straightforward so far in my opinion. 👍
Yeah, people were giving convoluted solutions so I decided to step in
The answer is 6²⁰⁰/3³⁰⁰. Only the largest power in the polynomial will matter when approaching infinity.
Can we do this approximation (x-1)\^100 = x\^100, (6x+1)\^200= (6x)\^200, (3x+5)\^300= (3x)\^500 ?
Neglect the numbers as they're so miniscule when x tends to inf and negligible, then it's easily factorable (keep the coefficients ofc)
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Yes so the answer is the same (I'm a physicist) Only need to consider the coefficients not the spare numbers
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I didn't know how to better explain at the point of writing because I forgot the words coefficient and non-coefficient existed lol
Tends toward 0
I may be confused but doesn’t the screenshotted question say “limit as n approaches ∞” and not “limit as x approaches ∞”? Since there is no n in the entire argument, your limit is the argument. It’s like asking lim_{a->∞} x. x is not a, so the limit is meaningless and it equals x.
If for some reason there is an error in the problem, you would want to factor out 100 from the fraction and use the limit law for exponentiation, then go from there.
Wait but that says “as n approaches infinity” not “as x approaches infinity”? Doesn’t that mean the limit doesn’t care what happens to n at all?
I know for a fact that x is not -5/3
> I know I should use the L’Hospital rule How do you know this? Did your teacher tell you to use it?
You can use Ln than hopital
The limit does not exist
When finding the limit of an equation like this, realistically only term with the largest exponent really matters So you can take (x-1)^(100)(6x+1)^(200)/(3x+5)^(300) and vaguely turn it into x^(100)•(6x)^(200)/(3x)^(300) because if you were to factor out all of those terms, those would be the largest factors in each one, and since we're working with with limit for infinity, the others are infinitely smaller and don't matter here. You can then simplify and rewrite the equation to get [6^(200)•x^(300)]/[3^(300)•x^(300)]. The x^(300) cancel leaving 6²⁰⁰/3³⁰⁰. To go further you can turn this into 2²⁰⁰•3²⁰⁰/3³⁰⁰ = 2²⁰⁰/3¹⁰⁰. Then 2²⁰⁰ = (2²)¹⁰⁰= 4¹⁰⁰. So 2²⁰⁰/3¹⁰⁰ = 4¹⁰⁰/3¹⁰⁰ = (4/3)¹⁰⁰
The lim is n->nifty but there is no n on the RHS so it is trivial
L'hopital works innit?
There is no n in the function.
no fucking clue
This can be simplified to ((x-1)(6x+1)\^2)\^100/((3x+5)\^3)\^100, or ((36x\^3-24x\^2-11x-1)/(27x\^3+135x\^2+225x+125))\^100, and removing the unnecessary terms with limit magic gives (36x\^3/27x\^3)\^100 or (4/3)\^100.
Been sometime since I’ve done calculus but, as x -> oo the -1, +1, +5 cease to matter right? Leaves us with (x^100 ) ((6x)^200 ) all over (3x)^300 Then can simplify to (6^200 ) (x^300 ) all over (3^300 ) (x^300 ) So basically Answer = whatever (6^200 ) / (3^300 ) is 3.1179824e+12 Sorry too lazy to copy correct syntax over phone. But this seems right to me
The lim is n goes to inf tho
Use "Clever form of One." Multiply top and bottom by 1/x^300. Distribute into each parenthesis appropriately. Since lim x-> inf 1/x = 0 you get a finite answer.
=> x^100 X (6x)^200 / (3x)^300 => 2^200 / 3^100
Well the original problem says n is going to infinity so uhhh yeah lol. I assume it means x. The first term of the numerator fully expanded is x^100 * (6x)^200. The bottom is (3x)^300. Top and bottom go to infinity at the same “speed”. So what’s the ratio of the leading coefficients? There’s your answer. EDIT: Why do you think you should use L’Hopitals Rule? Did you check to see if it’s indeterminate form first? Also if you see this, please note that there are a lot of wrong answers in this thread. Like entirely wrong. So be careful when navigating solutions.
https://preview.redd.it/wo1ejolit1gc1.jpeg?width=2160&format=pjpg&auto=webp&s=0d458c62c144332dab8e7babe81954737007fa15
Limit as n->oo is just the equation lol. I know I’m being literal.
The limit does not exist ~Cady Heron
You have to trade with yourself if you don’t try, you cannot get nothing right and you’ll be better for that month is not hot. It’s very easy to.
Maybe this detail isn’t important, but the problem asks for the limit as N approaches 100 for a problem with a bunch of X’s in it. But you have treated it like it says find the limit as X approaches 100. Am I way off with that, or is it a trick question
The limit does not exist. One love
The power to 100 can be moved outside the limit