Question

Find the limit of f(x)=3/x^4-1 as x approaches 1. The answer cannot be undefined.

Answer 1

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.

Answer 2

is f(x)=\[\frac{ 3 }{x ^{4}-1 }\]

Answer 3

??

Answer 4

Yes, that's right.

Answer 5

ohk
so factorize it as \[(x^{2}+1)(x-1)(x+1)\]

Answer 6

but in the numerator you must have a factor of x-1 else f(x) is undefined

Answer 7

I think you can use (x^3 -1/x) x

Answer 8

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.

Answer 9

can you post a quick screenshot of the material?

Answer 10

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

Answer 11

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

Answer 12

Alright, so in summary my answer is undefined no matter what?

Answer 13

hmm, no, is 0

Answer 14

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.

Answer 15

hmmm shoot I get something else from the graph :(

Answer 16

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)

Answer 17

i just referred that if there was a factor of x-1 in numerator we should have factorized it

Answer 18

it=x^4 -1 @DebbieG

Answer 19

I bet you entered 3/x^4-1 for the graph,, not 3/(x^4-1)... maybe?

Answer 20

Ok sorry but I'm still not clear on the answer and how to get there.

Answer 21

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

Answer 22

ok, wait... isn't the function we're talking about:

\[\Large f(x)=\frac{ 3 }{ x^4-1 }\]
Or no??

Answer 23

yes

Answer 24

Yes

Answer 25

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?

Answer 26

so yeah,.... if I stick rationals on \(\bf x^4\) it makes the fracton bigger, thus a bigger number off off to \(\bf +\infty \ and \ -\infty\)

Answer 27

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.