yueyue
  • yueyue
Find the limit of f(x)=3/x^4-1 as x approaches 1. The answer cannot be undefined.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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wolfe8
  • wolfe8
What you would want to do is factorize the denominator such that you can get 2 or more different fractions where one of them will not have 0 as the denominator when you substitute x.
anonymous
  • anonymous
is f(x)=\[\frac{ 3 }{x ^{4}-1 }\]
anonymous
  • anonymous
??

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yueyue
  • yueyue
Yes, that's right.
anonymous
  • anonymous
ohk so factorize it as \[(x^{2}+1)(x-1)(x+1)\]
anonymous
  • anonymous
but in the numerator you must have a factor of x-1 else f(x) is undefined
wolfe8
  • wolfe8
I think you can use (x^3 -1/x) x
yueyue
  • yueyue
Well the original problem asks me to find f(g(x)), where f(x) = 3/x-1 and g(x)= x^4, so there's not a factor of x-1 originally.
jdoe0001
  • jdoe0001
can you post a quick screenshot of the material?
yueyue
  • yueyue
anonymous
  • anonymous
it can be noted that if limiting value tends to infinity ans. is still valid i.e at neighborhood of x-->1 f(x) tends to assume very large values.. but its DISCONTINUOS at x=1
jdoe0001
  • jdoe0001
hmm... well, yes, that's correct, so if we stick a positive value to \(\bf x^4\) it will spit out a positive one, if we stick negative ones, it will also spit out positive ones the bigger the value, the greater the denominator, the smaller the fraction if the fraction gets smaller, then that means is approaching 0 that's is true for both one-sided limits though as hantenks correctly said, at 1 is undefined, but doesn't matter, the limit is just an approachment point
yueyue
  • yueyue
Alright, so in summary my answer is undefined no matter what?
jdoe0001
  • jdoe0001
hmm, no, is 0
DebbieG
  • DebbieG
Sorry... I don't understand, why do we want to factor \(x^4-1\)? what is the purpose of that? The limit as x->1+ is infinity, and limit as x->1- is -infinity.
jdoe0001
  • jdoe0001
hmmm shoot I get something else from the graph :(
DebbieG
  • DebbieG
There is a vertical asmptote at x=1. Since the den'r is non-zero, it's just a straightforward one-sided limit, isn't it? (maybe I'm missing something.. won't be the first time...lol)
anonymous
  • anonymous
i just referred that if there was a factor of x-1 in numerator we should have factorized it
anonymous
  • anonymous
it=x^4 -1 @DebbieG
DebbieG
  • DebbieG
I bet you entered 3/x^4-1 for the graph,, not 3/(x^4-1)... maybe?
yueyue
  • yueyue
Ok sorry but I'm still not clear on the answer and how to get there.
jdoe0001
  • jdoe0001
ohh, I see my mistake.... I was... anyhow, yes.. hhehe the graph is fine, it was just that I was using bigger integers for "x", thus moving away from 1
DebbieG
  • DebbieG
ok, wait... isn't the function we're talking about: \[\Large f(x)=\frac{ 3 }{ x^4-1 }\] Or no??
anonymous
  • anonymous
yes
yueyue
  • yueyue
Yes
DebbieG
  • DebbieG
OK, limits at x=1 differ from the left and from the right. Go to +inf from the right, -inf from the left. Hence, limit does not exist, but the 1-sided limits to, but are infinite. Right?
jdoe0001
  • jdoe0001
so yeah,.... if I stick rationals on \(\bf x^4\) it makes the fracton bigger, thus a bigger number off off to \(\bf +\infty \ and \ -\infty\)
wolfe8
  • wolfe8
Here is my solution although it's kinda radical:\[(x + 1)^{2}(x - 1)^{2}-x ^{3}+x ^{2}+2x\] as the denominator. Then substituting x = 1 in the equation will have a definite answer. I haven't checked if I expanded the brackets correctly though.
DebbieG
  • DebbieG
|dw:1378755920939:dw|
yueyue
  • yueyue
Since I have to show my work. Would I just solve the and leave it undefined? Then just explain?
DebbieG
  • DebbieG
@hantenks approaching 1 from the right, x^4 >1 so the den'r is positive. it will shoot to +inf, not -inf. Opposite on the other side of 1.
jdoe0001
  • jdoe0001
hmmmm
wolfe8
  • wolfe8
And yes it doesn't have a finite value, but we are looking at limits. So we are looking at the value it seems to be approaching.
DebbieG
  • DebbieG
I'm not sure I would say "undefined"... I would say the limit "does not exist". By the definition of limit, if the limits from left and right are not the same, the limit does not exist.
anonymous
  • anonymous
@DebbieG you rock
DebbieG
  • DebbieG
But limit DNE at x=a if limit from right and limit from left are not equal. The one-sided limits exist, but the question didn't ask about those, just the general limit as x->1, which DNE.
DebbieG
  • DebbieG
lol I don't know about that but thanks @hantenks
yueyue
  • yueyue
Ok given that the limit does not exist. Then what would I do for part C of the question? It asks me to find all values of x that are discontinuous.
DebbieG
  • DebbieG
Well, x=1 is clearly one of them. There is at least one other. Where else is the function undefined?
DebbieG
  • DebbieG
(This is where the factored form of the den'r, although not strictly necessary, could be helpful. But the even-root property would also suffice.)
yueyue
  • yueyue
I sincerely have no idea. It is needless to say, I am lost but thank you all for your help.
jdoe0001
  • jdoe0001
\(\bf \cfrac{3}{x^4-1} \implies \cfrac{3}{(x^2+1)(x-1)(x+1)}\)
DebbieG
  • DebbieG
It's a rational function. A rational function is continuous for all values of x, EXCEPT those that make the den'r=0 so if you solve: \[\Large x^4-1=0\] ... you will have all the x's for which the function is discontinuous.
DebbieG
  • DebbieG
You can use the factoring as @jdoe0001 demonstrated, or just apply the even root property and take the 4th root of both sides (luckily, easy to do here: \[\Large x^4=1\] \[\Large x=\pm \sqrt[4]{1}\]
yueyue
  • yueyue
Alright I got that. Thank you @DebbieG and @jdoe0001 .
jdoe0001
  • jdoe0001
yw

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