F(0)=0
xcos(1/x) when x=\0
Check differential at x=0
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neon quest
native jewelBOT
F(0)=0
xcos(1/x) when x=\0
Check differential at x=0
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neon quest
So here i did this
hcos(1/h)/h
cos(1/h)=cos(-1/h)
And both are equal
So it will be differentiable at x=0
fathom forge
So it will be differentiable at x=0
$\lim_{h\to 0}\cos(1/h)$ DNE
supple condorBOT
Omegabet_
fathom forge
idk where the evenness came up
neon quest
,w sin (1/x) tends to 0
fathom forge
sin(1/x) also is DNE as x goes to 0...
supple condorBOT
Requested by arjunn5589
fathom forge
obvious from the fact cos(t) as t goes to +-inf DNE
But anyway, like the question says you just check 1st principles and whether that holds or not at 0
hence checking whether $\lim_{h\to 0}\frac{h\cos(1/h)-0}{h}=\lim_{h\to 0}\cos(1/h)$ exists.
supple condorBOT
Omegabet_
fathom forge
Because cos can handle negative inside
a function $f$ is differentiable at $a$ iff $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists and is finite
supple condorBOT
Omegabet_
fathom forge
Not... its domain can have negative numbers
xcos(1/x) is differenatiable at 0 iff the limit as h goes to 0 of cos(1/h) exists.
Said limit clearly DNE, so xcos(1/x) is clearly not differentiable at 0.
fathom forge
Ok, let's try it this way
neon quest
And f'(0)=hcos(1/h)-0/h
cos (1/h) and cos(-1/h) will be same
fathom forge
what's $\lim_{x\to\infty}\cos(x)$?
supple condorBOT
Omegabet_
fathom forge
and $\lim_{x\to-\infty}\cos(x)$?
supple condorBOT
Omegabet_
neon quest
DNE
fathom forge
So now set $x:=1/h$ in $\lim_{h\to 0}\cos(1/h)$, what does that sub do?
supple condorBOT
Omegabet_
neon quest
Ahh it will be DNE surely
fathom forge
yes, you get $\lim_{h\to 0}\cos(1/h)=\lim_{x\to\pm\infty}\cos(x)$
neon quest
But it is between [-1 1] oscillation
supple condorBOT
Omegabet_
fathom forge
yeah, and?
neon quest
Ohh so limit exist but differntiation no
fathom forge
Ok, try reading this time
neon quest
Sure
neon quest
Yes true statement
fathom forge
applying that to $f(x)=x\cos(1/x)$ and $a=0$... you want to check whether $\lim_{h\to 0}\frac{h\cos(1/h)-0}{h}$ exists
supple condorBOT
Omegabet_
fathom forge
that limit is clearly, from day 1 algebra, equal to $\lim_{h\to 0}\cos(1/h)$
supple condorBOT
Omegabet_
fathom forge
so xcos(1/x) is differentiable at 0 iff the limit as h goes to 0 of cos(1/h) exists.
per the literal definition verbatim
neon quest
I am saying the limit x=0 exist for the function but differential no
fathom forge
LEARN TO READ, THE LIMIT DNE AS YOU YOURSELF SAID
neon quest
**LEARN TO READ, THE LIMIT DNE AS YOU YOURSELF SAID**
Limit for differentiation doesn't
fathom forge
yes
neon quest
Limit for limit exist
fathom forge
"limit for limit" is nonsense
but checking continuity wasnt the question so idk why you care
neon quest
Limit h tends to 0+
h×[-1 1] =0
Limit h tends to 0-
-h×[-1 1] =0
So the limit at x=0 exist
neon quest
I don't know in our exams we first check the limit exists or not
fathom forge
yes, you're checking whether the 1st principles limit exists....
cause that's the question
neon quest
Yes.
fathom forge
If you opted to attempt a question you never read, then I wish you luck cause your brain is backwards lol
But at this rate you're just trolling and seems like you're purposefully being daft