Here's the question you clicked on:
recon14193
limit evaluation
\[\lim_{x \rightarrow -4}\frac{ \sqrt{x^9+9}-5 }{ x+4 }\]
Try rationalizing the top by multiply the numerator and denominator by the top's conjugate
I worked it out to \[\frac{ x^9-16 }{ x+4(\sqrt{x^9+9}+5) }\]
I dont know what I would expand the top to
and i think you mean to have parenthesis around the (x+4) too on bottom
try 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
I dont know how I would factor the top though and also with it being ^9 it's still a negative square root
do you know how do synthetic division or long division?
I should but I dont remember
ok 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
if 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
the 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
how do you get the limit exists?
This is part of a review and when I put DNE, I was told I was incorrect
This 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
you 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)
http://www.wolframalpha.com/input/?i=y%3D%28sqrt%28x%5E9%2B9%29-5%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
i 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
as you can also tell from the graph nothing exists around x=-4
the limit does not exist we have determine this algebraically, graphically, and numerically
\[\lim_{x \rightarrow -4}\frac{\sqrt{x^9+9}-5}{x+4}\] just to be certain this was the problem right?
so what is needed to convince you this limit does not exist? like which explanation does not make sense?
It'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
we can try to determine where the written went wrong in writing the test
give 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
could I ask you one other problem real quick first
so if they are saying the limit exist then that means they want some (x+4)'s to cancel... so one second
http://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
I got the -2 but not why -1 also
do me a favor while i look at that try to answer that same question as if you were answering \[\lim_{x \rightarrow -4}\frac{\sqrt{x^2+9}-5}{x+4}\]
i think you might get it right if you pretend that is the question
let me know and if you get it wrong i'm out of guesses of what they meant
ok you have a fraction inside that fraction
that one fraction does not exist at x=-1
and then after that of course you already knew to find when 1+1/(x+1) is not 0 which is when x=-2
for 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
alright so for \[\frac{ \sqrt{x^2+9}-5 }{ x+4 }\] I get\[\frac{ (x-4) }{ \sqrt{x^2+9}+5 }\] which still gives a negative root when I plug in -4
so 1/(1+1/0)) is still not a number
so yeah never mind it gives 0
ok you plugged in -4 right?
\[\frac{-4-4}{\sqrt{(-4)^2+9}+5}\] I think you are getting tired of this problem and you aren't going through the arithmetic slowly enough
and can you do the bottom?
\[\sqrt{16+9}=\sqrt{25}....... 5+5=10\] so 8/10 or 4/5
don't forget the negative sign from the top
so -4/5 tell me if that is what they were looking for or not
I don't know because it doesn't show me the correct answer, but it is something
you should let your teacher know of the mistake though
most teachers (or some teachers) are amazed by students who can find there errors
I haven't met them yet, and I don't head to school until another week. What would be the most respectable non-annoying way to present it?
Definitely 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.
Or dear professor if you are feeling more formal.
alright should I provide an explanation of my work about how I could not find the limit algebraically, graphically, or numerically for the original problem
I 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)
alright thank you for all you help
Np. 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.
Good luck @recon14193