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Differentiability of $g(x)sin(1/x)$ at $x=0$.
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I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:
Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
$$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
is differentiable at $x=0$.
My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
begin{align*}
lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
\ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
end{align*}
Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
$$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
Hence, $f'(0)=0$.
I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.
real-analysis proof-verification
asked 23 hours ago
高田航
1,274318
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up vote
0
down vote
favorite
I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:
Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
$$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
is differentiable at $x=0$.
My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
begin{align*}
lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
\ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
end{align*}
Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
$$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
Hence, $f'(0)=0$.
I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.
real-analysis proof-verification
asked 23 hours ago
高田航
1,274318
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:
Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
$$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
is differentiable at $x=0$.
My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
begin{align*}
lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
\ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
end{align*}
Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
$$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
Hence, $f'(0)=0$.
I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.
real-analysis proof-verification
asked 23 hours ago
高田航
1,274318
I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:
Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
$$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
is differentiable at $x=0$.
My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
begin{align*}
lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
\ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
end{align*}
Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
$$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
Hence, $f'(0)=0$.
I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.
real-analysis proof-verification
real-analysis proof-verification
asked 23 hours ago
高田航
1,274318
asked 23 hours ago
高田航
1,274318
asked 23 hours ago
高田航
1,274318
asked 23 hours ago
高田航
1,274318
asked 23 hours ago
高田航
1,274318
1,274318
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
The limit $lim_{hto 0}sin(1/h)$ does not exist !
But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.
answered 23 hours ago
Fred
41.9k1642
add a comment |
up vote
2
down vote
Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.
It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$
answered 23 hours ago
José Carlos Santos
139k18111203
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up vote
1
down vote
If I am correct then the question mention's about
$$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$
So I would use the following:
$$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$
$$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$
Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.
This shortened the expression to
$$ lim_{h rightarrow 0} g(h) $$
Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.
answered 23 hours ago
Lakshya Sinha
464
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up vote
0
down vote
By Taylor's theorem:
$$g(x)=g(0)+g'(0)x+r(x)x$$
$$g(x)=xr(x)$$
So
$$f(x)=xr(x)sinleft(frac{1}{x}right)$$
And
$$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.
answered 23 hours ago
Botond
4,9632732
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The limit $lim_{hto 0}sin(1/h)$ does not exist !
But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.
answered 23 hours ago
Fred
41.9k1642
add a comment |
up vote
4
down vote
accepted
The limit $lim_{hto 0}sin(1/h)$ does not exist !
But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.
answered 23 hours ago
Fred
41.9k1642
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The limit $lim_{hto 0}sin(1/h)$ does not exist !
But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.
answered 23 hours ago
Fred
41.9k1642
The limit $lim_{hto 0}sin(1/h)$ does not exist !
But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.
answered 23 hours ago
Fred
41.9k1642
answered 23 hours ago
Fred
41.9k1642
answered 23 hours ago
Fred
41.9k1642
answered 23 hours ago
Fred
41.9k1642
41.9k1642
add a comment |
add a comment |
up vote
2
down vote
Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.
It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$
answered 23 hours ago
José Carlos Santos
139k18111203
add a comment |
up vote
2
down vote
Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.
It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$
answered 23 hours ago
José Carlos Santos
139k18111203
add a comment |
up vote
2
down vote
up vote
2
down vote
Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.
It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$
answered 23 hours ago
José Carlos Santos
139k18111203
Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.
It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$
answered 23 hours ago
José Carlos Santos
139k18111203
answered 23 hours ago
José Carlos Santos
139k18111203
answered 23 hours ago
José Carlos Santos
139k18111203
answered 23 hours ago
José Carlos Santos
139k18111203
139k18111203
add a comment |
add a comment |
up vote
1
down vote
If I am correct then the question mention's about
$$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$
So I would use the following:
$$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$
$$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$
Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.
This shortened the expression to
$$ lim_{h rightarrow 0} g(h) $$
Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.
answered 23 hours ago
Lakshya Sinha
464
add a comment |
up vote
1
down vote
If I am correct then the question mention's about
$$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$
So I would use the following:
$$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$
$$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$
Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.
This shortened the expression to
$$ lim_{h rightarrow 0} g(h) $$
Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.
answered 23 hours ago
Lakshya Sinha
464
add a comment |
up vote
1
down vote
up vote
1
down vote
If I am correct then the question mention's about
$$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$
So I would use the following:
$$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$
$$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$
Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.
This shortened the expression to
$$ lim_{h rightarrow 0} g(h) $$
Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.
answered 23 hours ago
Lakshya Sinha
464
If I am correct then the question mention's about
$$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$
So I would use the following:
$$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$
$$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$
Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.
This shortened the expression to
$$ lim_{h rightarrow 0} g(h) $$
Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.
answered 23 hours ago
Lakshya Sinha
464
answered 23 hours ago
Lakshya Sinha
464
answered 23 hours ago
Lakshya Sinha
464
answered 23 hours ago
Lakshya Sinha
464
464
add a comment |
add a comment |
up vote
0
down vote
By Taylor's theorem:
$$g(x)=g(0)+g'(0)x+r(x)x$$
$$g(x)=xr(x)$$
So
$$f(x)=xr(x)sinleft(frac{1}{x}right)$$
And
$$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.
answered 23 hours ago
Botond
4,9632732
add a comment |
up vote
0
down vote
By Taylor's theorem:
$$g(x)=g(0)+g'(0)x+r(x)x$$
$$g(x)=xr(x)$$
So
$$f(x)=xr(x)sinleft(frac{1}{x}right)$$
And
$$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.
answered 23 hours ago
Botond
4,9632732
add a comment |
up vote
0
down vote
up vote
0
down vote
By Taylor's theorem:
$$g(x)=g(0)+g'(0)x+r(x)x$$
$$g(x)=xr(x)$$
So
$$f(x)=xr(x)sinleft(frac{1}{x}right)$$
And
$$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.
answered 23 hours ago
Botond
4,9632732
By Taylor's theorem:
$$g(x)=g(0)+g'(0)x+r(x)x$$
$$g(x)=xr(x)$$
So
$$f(x)=xr(x)sinleft(frac{1}{x}right)$$
And
$$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.
answered 23 hours ago
Botond
4,9632732
answered 23 hours ago
Botond
4,9632732
answered 23 hours ago
Botond
4,9632732
answered 23 hours ago
Botond
4,9632732
4,9632732
add a comment |
add a comment |
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