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recon14193
 8 months ago
limit evaluation
recon14193
 8 months ago
limit evaluation

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recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 4}\frac{ \sqrt{x^9+9}5 }{ x+4 }\]

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1Try rationalizing the top by multiply the numerator and denominator by the top's conjugate

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I worked it out to \[\frac{ x^916 }{ x+4(\sqrt{x^9+9}+5) }\]

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I dont know what I would expand the top to

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1and i think you mean to have parenthesis around the (x+4) too on bottom

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1try factoring the top to see if there is a factor of x+4 in it so you can get rid of what would make the bottom 0

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I dont know how I would factor the top though and also with it being ^9 it's still a negative square root

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1do you know how do synthetic division or long division?

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I should but I dont remember

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1ok well if you did know how and you determine there was no remainder then the limit does exist but if there is a remainder then the limit will not exist but i guess you can also determine this from just pluggin in 4 on both top and bottom if you have 0/0 the limit will exist if you don't then it won't

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1if the limit does exist then you of course must find it but if the limit doesn't exist, then you are done and dne is your answer

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0the limit does exist I know that, and I went back to my algebra 2 notes and doing synthetic division (x+4) is not a factor

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1how do you get the limit exists?

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0This is part of a review and when I put DNE, I was told I was incorrect

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1This question doesn't have a real limit had the function only exist for values where x^9+9 is greater than or equal to 0 anyhow x^9 has to be greater to 0 which means x cannot be negative because if it makes since to say if x is negative then x^9 is definitely more and most negative

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1you can even see a numerical approach will also show nothing exists to the left and right of 4 or at 4 (even though we aren't really concerned what happens at 4)

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=y%3D%28sqrt%28x%5E9%2B9%295%29%29%2F%28x%2B4%29+real of course there are some negative values the function does exist for after all x^9+9 has to be greater than or equal to 0

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1i misphrased it a little above when i just said x^9 has to be greater than or equal to 0 only small negative values will work anyhow is what i mean

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1as you can also tell from the graph nothing exists around x=4

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1the limit does not exist we have determine this algebraically, graphically, and numerically

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow 4}\frac{\sqrt{x^9+9}5}{x+4}\] just to be certain this was the problem right?

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1so what is needed to convince you this limit does not exist? like which explanation does not make sense?

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0It's not that none of the explanations don't make sense. I came to the same conclusion working it myself. It's just that when I plugged in my answer what the test told me was that, that answer is incorrect

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1we can try to determine where the written went wrong in writing the test

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1give me a second and let me see if i can find out where they went wrong in writing the problem because i bet you that is what happened

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0could I ask you one other problem real quick first

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1so if they are saying the limit exist then that means they want some (x+4)'s to cancel... so one second

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%28x%2B1%29%2F%281%2B%281%2F%28x%2B1%29%29%29 For the domain why is it x=/=2 and x=/=1

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I got the 2 but not why 1 also

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1do me a favor while i look at that try to answer that same question as if you were answering \[\lim_{x \rightarrow 4}\frac{\sqrt{x^2+9}5}{x+4}\]

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1i think you might get it right if you pretend that is the question

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1let me know and if you get it wrong i'm out of guesses of what they meant

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1ok you have a fraction inside that fraction

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1that one fraction does not exist at x=1

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1and then after that of course you already knew to find when 1+1/(x+1) is not 0 which is when x=2

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1for example when pluggin in 1 we get \[\frac{1}{1+\frac{1}{1+1}}=\frac{1}{1+\frac{1}{0}} \] but 1/0 is not a number

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0alright so for \[\frac{ \sqrt{x^2+9}5 }{ x+4 }\] I get\[\frac{ (x4) }{ \sqrt{x^2+9}+5 }\] which still gives a negative root when I plug in 4

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1so 1/(1+1/0)) is still not a number

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0so yeah never mind it gives 0

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1ok you plugged in 4 right?

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1\[\frac{44}{\sqrt{(4)^2+9}+5}\] I think you are getting tired of this problem and you aren't going through the arithmetic slowly enough

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1and can you do the bottom?

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0\[\sqrt{16+9}=\sqrt{25}....... 5+5=10\] so 8/10 or 4/5

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1don't forget the negative sign from the top

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1so 4/5 tell me if that is what they were looking for or not

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I don't know because it doesn't show me the correct answer, but it is something

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1you should let your teacher know of the mistake though

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1most teachers (or some teachers) are amazed by students who can find there errors

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0I haven't met them yet, and I don't head to school until another week. What would be the most respectable nonannoying way to present it?

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1Definitely don't rub in there face. I would just bring it up once if and when they respond. I think an email would be just fine. You can be like: Hey professor, As I was taking the exam, I came across the problem lim (x>4) (sqrt(x^9+9)5)/(x+4) and I said the limit does not exist but it was returned incorrect by the exam. I was thinking it was meant to be lim (x>4) (sqrt(x^2+9)5)/(x+4) where the answer would be 4/5. Please let me know if I'm correct or not. Sincerely, (you) But don't say she made a mistake. Just say the exam or whatever. What I'm saying is don't say YOU in the email. Try not to imply it is the teacher's fault even though it is. I think that is the best way.

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1Or dear professor if you are feeling more formal.

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0alright should I provide an explanation of my work about how I could not find the limit algebraically, graphically, or numerically for the original problem

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1I think you could say just nothing exists around 4 since x^9+9 has to be greater than or equal to 0. You know which means x^9>=9 But for x values less than 4 certainly do not satisfy the inequality x^9>=9. You could actually solve that inequality if you wanted to. x>=(9)^(1/9) but values really close to 4 are definitely not in the set [(9)^(1/9),inf)

recon14193
 8 months ago
Best ResponseYou've already chosen the best response.0alright thank you for all you help

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1Np. I don't have problem helping you with this one on your test mainly because it was written wrong. But I hope you intend to give the rest of the test your own knowledge.

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.1Good luck @recon14193
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