I want to replace the question mark with an expression that makes the following true
#Inequality problem
opal heron
near ibexBOT
I want to replace the question mark with an expression that makes the following ...
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winter bluffBOT
ASR is Mathematicoid
snow spruce
$$x^4 +\frac{1}{x}-2 = (x^4-1) + \frac{x-1}{x}$$
winter bluffBOT
aL
snow spruce
you have control over |x-1|
suppose |x-1| < 1/2, for example and bound the remaining quantities
$$=(x-1) \left( (x^2+1)(x+1) + \frac{1}{x} \right)$$
winter bluffBOT
aL
snow spruce
you can find a bound for the expression in the big parentheses
opal heron
The 1/x term is causing some trouble
opal heron
Am I supposed to find bounds for the quantitis in the parentheses as
|x-1|<1/2 --> |x|<3/2
Then |x^2+1|=x^2+1<9/4+1=13/4
snow spruce
that suffices
you don't have to make the bounds "optimal" so to speak
note that 1/2 < x < 3/2 < 2
so you can say
$$(x^2+1)(x+1) + \frac{1}{x} \leqslant 5\cdot 3 + 2$$
winter bluffBOT
aL
snow spruce
$$\delta \leqslant \min \left\lbrace \frac{1}{2},\frac{\varepsilon}{20} \right\rbrace$$
winter bluffBOT
aL
snow spruce
this is enough, for example
opal heron
Am I supposed to find bounds for the quantitis in the parentheses as
|x-1|<1/2 -...
But when I try to do it with |1/x| I cannot say anymore that since |x|<3/2 |1/x|<1/3/2, how to find bound for it
snow spruce
$$\frac{1}{2}< x \Leftrightarrow \frac{1}{x} < 2$$
winter bluffBOT
aL
snow spruce
if you invert, you have to use the lower bound
snow spruce
$$|x-1| < \frac{1}{2} \Leftrightarrow \frac{1}{2} < x < \frac{3}{2}$$
winter bluffBOT
aL
snow spruce
Didn't we assume that |x|<1/2
we certainly didn't
you don't want x to get close to 0
opal heron
x is already positive
Its because of assuming |x-1|<1/2 and in general, letting |x-1| smaller than a number <=1. Can't we do otherwise
opal heron
I mean a number larger than 1
snow spruce
no
opal heron
Why tho
tardy gazelleBOT
Congratulations @snow spruce, you have been awarded the <@&1257594408103317575> for being the most active user today.
snow spruce
for fuck's sake..
anyway, think about it like this
we are interested in the behaviour of the function around x=1
we don't care what happens with the function values for x that are far away from 1
that's why you start by restricting |x-1| < 1/5 or whatever you want
opal heron
we don't care what happens with the function values for x that are far away from...
That's the question. Ever since I have started the study of limits I am reading stuff like "It is of smaller epsilons (and perhaps deltas) which are of interest". Why?
snow spruce
what does a large epsilon reveal?
opal heron
what does a large epsilon reveal?
What, idk
snow spruce
me neither
opal heron
me neither
snow spruce
you want to prove lim x^2 +1 = 1 as x to 0, for example
if you say |x| < 1/2, then it is always true that |x^2+1-1| < 10
this is of no help to determine whether the limit is 1
on the other hand, I can make the epsilon smaller
if |x| < 1/2, then it is not necessarily true anymore that |x^2| < 1/16
in fact, if I let epsilon be any positive small number, am I still able to find an open interval around x=0 ?
no matter how small
if the answer is yes, then the limit is 1
opal heron
+close
tardy lagoonBOT
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tardy lagoonBOT
@opal heron