Mixing
Approximation of a differentiable function.
$begingroup$
If $Doverset{f}{rightarrow}mathbb{C}$ is differentiable in $x_0$ $(Dsubseteqmathbb{R})$, the closer one gets to $x_0$ the smaller the difference between the linear function $l_{x_{0}}(x)=f(x_0)+f'(x_0)(x-x_0)$ gets. The Approximation is defined as:
$$begin{align}
r_{x_{0}}(x) &:=f(x)-l_{x_{0}}(x)\
&=f(x)-f(x_0)-f'(x_0)(x-x_0) \
&=left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right)(x-x_0)
end{align}$$
How does one conclude that the left factor on the very Right side converges to $0$.
And how does one conclude by that that
$lim_{xrightarrow x_0}frac{r_{x_{0}}(x)}{x-x_0}=0$
real-analysis
edited Jan 19 at 21:13
BigbearZzz
8,88821652
asked Jan 19 at 19:54
RM777RM777
36912
$endgroup$
add a comment |
$begingroup$
If $Doverset{f}{rightarrow}mathbb{C}$ is differentiable in $x_0$ $(Dsubseteqmathbb{R})$, the closer one gets to $x_0$ the smaller the difference between the linear function $l_{x_{0}}(x)=f(x_0)+f'(x_0)(x-x_0)$ gets. The Approximation is defined as:
$$begin{align}
r_{x_{0}}(x) &:=f(x)-l_{x_{0}}(x)\
&=f(x)-f(x_0)-f'(x_0)(x-x_0) \
&=left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right)(x-x_0)
end{align}$$
How does one conclude that the left factor on the very Right side converges to $0$.
And how does one conclude by that that
$lim_{xrightarrow x_0}frac{r_{x_{0}}(x)}{x-x_0}=0$
real-analysis
edited Jan 19 at 21:13
BigbearZzz
8,88821652
asked Jan 19 at 19:54
RM777RM777
36912
$endgroup$
$begingroup$
the limit of the left factor is zero by definition of derivative at $x_0$
$endgroup$
– GReyes
Jan 19 at 20:13
$begingroup$
You divide by $(x-x_0)$ both sides and use the fact that that left factor converges to zero
$endgroup$
– GReyes
Jan 19 at 20:18
add a comment |
$begingroup$
If $Doverset{f}{rightarrow}mathbb{C}$ is differentiable in $x_0$ $(Dsubseteqmathbb{R})$, the closer one gets to $x_0$ the smaller the difference between the linear function $l_{x_{0}}(x)=f(x_0)+f'(x_0)(x-x_0)$ gets. The Approximation is defined as:
$$begin{align}
r_{x_{0}}(x) &:=f(x)-l_{x_{0}}(x)\
&=f(x)-f(x_0)-f'(x_0)(x-x_0) \
&=left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right)(x-x_0)
end{align}$$
How does one conclude that the left factor on the very Right side converges to $0$.
And how does one conclude by that that
$lim_{xrightarrow x_0}frac{r_{x_{0}}(x)}{x-x_0}=0$
real-analysis
edited Jan 19 at 21:13
BigbearZzz
8,88821652
asked Jan 19 at 19:54
RM777RM777
36912
$endgroup$
If $Doverset{f}{rightarrow}mathbb{C}$ is differentiable in $x_0$ $(Dsubseteqmathbb{R})$, the closer one gets to $x_0$ the smaller the difference between the linear function $l_{x_{0}}(x)=f(x_0)+f'(x_0)(x-x_0)$ gets. The Approximation is defined as:
$$begin{align}
r_{x_{0}}(x) &:=f(x)-l_{x_{0}}(x)\
&=f(x)-f(x_0)-f'(x_0)(x-x_0) \
&=left(frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)right)(x-x_0)
end{align}$$
How does one conclude that the left factor on the very Right side converges to $0$.
And how does one conclude by that that
$lim_{xrightarrow x_0}frac{r_{x_{0}}(x)}{x-x_0}=0$
real-analysis
real-analysis
edited Jan 19 at 21:13
BigbearZzz
8,88821652
asked Jan 19 at 19:54
RM777RM777
36912
edited Jan 19 at 21:13
BigbearZzz
8,88821652
asked Jan 19 at 19:54
RM777RM777
36912
edited Jan 19 at 21:13
BigbearZzz
8,88821652
edited Jan 19 at 21:13
BigbearZzz
8,88821652
edited Jan 19 at 21:13
BigbearZzz
8,88821652
8,88821652
asked Jan 19 at 19:54
RM777RM777
36912
asked Jan 19 at 19:54
RM777RM777
36912
asked Jan 19 at 19:54
RM777RM777
36912
36912
$begingroup$
the limit of the left factor is zero by definition of derivative at $x_0$
$endgroup$
– GReyes
Jan 19 at 20:13
$begingroup$
You divide by $(x-x_0)$ both sides and use the fact that that left factor converges to zero
$endgroup$
– GReyes
Jan 19 at 20:18
add a comment |
$begingroup$
the limit of the left factor is zero by definition of derivative at $x_0$
$endgroup$
– GReyes
Jan 19 at 20:13
$begingroup$
You divide by $(x-x_0)$ both sides and use the fact that that left factor converges to zero
$endgroup$
– GReyes
Jan 19 at 20:18
$begingroup$
the limit of the left factor is zero by definition of derivative at $x_0$
$endgroup$
– GReyes
Jan 19 at 20:13
$begingroup$
the limit of the left factor is zero by definition of derivative at $x_0$
$endgroup$
– GReyes
Jan 19 at 20:13
$begingroup$
You divide by $(x-x_0)$ both sides and use the fact that that left factor converges to zero
$endgroup$
– GReyes
Jan 19 at 20:18
$begingroup$
You divide by $(x-x_0)$ both sides and use the fact that that left factor converges to zero
$endgroup$
– GReyes
Jan 19 at 20:18
add a comment |
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$begingroup$
the limit of the left factor is zero by definition of derivative at $x_0$
$endgroup$
– GReyes
Jan 19 at 20:13
$begingroup$
You divide by $(x-x_0)$ both sides and use the fact that that left factor converges to zero
$endgroup$
– GReyes
Jan 19 at 20:18