proving limit law by definition

**transgalactic** · August 2nd 2011, 03:18 AM

there is $lim_{x->x_{0}}g(x)=y_{0}$ , $f(x)$ is continues in $y_{0}$
point. i need to prove that $lim_{x->x_{0}}f\circ g(x)=f(lim_{x->x_{0}}g(x))=f(y_{0})$
by epsilon delta definition.
i need to follow only my books prove.help me understand it.
the books proof:
in order to prove $lim_{x->x_{0}}f\circ g(x)=f(lim_{x->x_{0}}g(x))=f(y_{0})$
we will use the definition that for $N_{\varepsilon}^{*}(f(y_{0}))$
surrounding we have $N_{\delta}^{*}(x)$ surrounding for which every
$x\in N_{\delta}^{*}(x)$ $f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$ .
first step:
$f(x)$ is continues in $y_{0}$ point. so $lim_{x->x_{0}}f(x)=f(y_{0})$
and for $N_{\varepsilon}^{*}(f(y_{0}))$ surrounding we have $N_{\delta}^{*}(x)$
surrounding for which every
$x\in N_{\delta}^{*}(x)$ $f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$ .
second step:
it is given that $lim_{x->x_{0}}g(x)=y_{0}$ then for $x\in N_{\delta_{1}}^{*}(x)$
$g(x)\in N_{\varepsilon}^{*}(y_{0})$
so $f((g(x)))$ $\in N_{\varepsilon}^{*}(f(y_{0}))$ .
cant understand this last step why $f((g(x)))$ $\in N_{\varepsilon}^{*}(f(y_{0}))$
is the resolt of the last two steps??

**HallsofIvy** · August 2nd 2011, 03:45 AM

You first step concludes that if $x\in N_{\delta}^{*}(x)$ then $f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$ .

Your second step concludes that if $x\in N^*_\delta$ then $g(x)\in N_{\varepsilon}^{*}(y_{0})$

The second step says that g(x) satisfies the condition in the first step so you replace "x" in the first step by "g(x)".

**transgalactic** · August 2nd 2011, 03:52 AM

but we have different deltas in each limit
the first is delta1 the other is just delta.

and g doesnt satisfies the first one
because its a different surroundings

cant see wht we replace x with g(x)
?

why g(x) suttisfies the first step?

**Plato** · August 2nd 2011, 04:24 AM

Originally Posted by transgalactic

but we have different deltas in each limit
the first is delta1 the other is just delta.
and g doesnt satisfies the first one
because its a different surroundings
cant see wht we replace x with g(x)?
why g(x) suttisfies the first step?

To be truthful, the book took some liberties with that proof.
It would have been better to say, there is $\delta_1>0$ for the continuity and there is $\delta_2>0$ for the sequential limit.
Then let $\delta=\min\{\delta_1,~\delta_2\}$ .
That way it insures both conditions hold.

**transgalactic** · August 2nd 2011, 04:26 AM

ok that is true if i choose minimal delta like you said.
why g(x) suttisfies the first condition
?

**Plato** · August 2nd 2011, 04:33 AM

Originally Posted by transgalactic

ok that is true if i choose minimal delta like you said. why g(x) suttisfies the first condition?

Because if $\delta\le\delta_2$ then $N_\delta ^* (x_0 ) \subseteq N_{\delta _2 }^* (x_0 )$ .

**transgalactic** · August 2nd 2011, 04:42 AM

Originally Posted by HallsofIvy

You first step concludes that if $x\in N_{\delta}^{*}(x)$ then $f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$ .

Your second step concludes that if $x\in N^*_\delta$ then $g(x)\in N_{\varepsilon}^{*}(y_{0})$

The second step says that g(x) satisfies the condition in the first step so you replace "x" in the first step by "g(x)".

the minimal delta thing only says that there is a smaller surrounding for which
$f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$
and
$g(x)\in N_{\varepsilon}^{*}(y_{0})$

how from it we cant conlude that we can put g(x) instead of x

?

**Plato** · August 2nd 2011, 06:48 AM

Originally Posted by transgalactic

the minimal delta thing only says that there is a smaller surrounding for which
$f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$
and
$\color{red}g(x)\in N_{\varepsilon}^{*}(y_{0})$
how from it we cant conlude that we can put g(x) instead of x

You don't want $g(x)\in N_{\varepsilon}^{*}(y_{0})$ !

You want $f(g(x_0))\in N_{\varepsilon}^{*}(y_{0})$

**transgalactic** · August 2nd 2011, 06:53 AM

yes i do

i just dont know how to get to this resolt from those two expression i am givem by taking the minimal delta

?

**Plato** · August 2nd 2011, 07:29 AM

Originally Posted by transgalactic

i just dont know how to get to this resolt from those two expression i am givem by taking the minimal delta

Given that $f$ is continuous at $y_0$ and $\lim _{x \to x_0 } g(x) = y_0$

Suppose that $\varepsilon > 0$ .

From continuity $\left( {\exists \delta _1 >0} \right)\left[ {x \in N_{\delta _1 }^* (y_0 )\, \Rightarrow \,f(x) \in N_\varepsilon ^* \left( {f(y_0 )} \right)} \right]$

From the limit, $\left( {\exists \delta _2 > 0} \right)\left[ {x \in N_{\delta _2 }^* (x_0 )\, \Rightarrow \,g(x) \in N_{\delta _1 }^* \left( {y_0 } \right)} \right]$ .

Let $\delta = \min \left\{ {\delta _1 ,\delta _2 } \right\}$ .

Then $\left[ {x \in N_\delta ^* (x_0 )\, \Rightarrow g(x) \in N_{\delta _1 }^* (y_0 )\, \Rightarrow \,\,f\left( {g(x)} \right) \in N_\varepsilon ^* \left( {f(y_0 )} \right)} \right]$

**transgalactic** · August 2nd 2011, 08:03 AM

ok i start to understand the idea
how do i know that if we put g(x) instead of x
i get an input of 'x' in f(x) for which this 'x' is inside the section bounded by delta1
?
maybe this 'x' is out side the section bounded by delta1.
we dont know what is g(x) we can put some x which is bounded by the minimal delta
and get alot bigger number

for example
g(x)=1/x
a=0.001
g(a)=1000
and get fg(a)=f(1000)
and this 1000 could be higher the the minimal delta
??

**Plato** · August 2nd 2011, 08:34 AM

This is my last post to this thread.
(NOTATION: $a\approx b$ means that $a\ne b\text{ and }a\text{ is ‘close to’ }b$

From continuity if $z\approx y_0$ then $f(z)\approx f(y_0)$ .

From the limit, if $w\approx x_0$ then $g(w)\approx y_0$

Now if $g(w)\approx y_0$ that means $f(g(w))\approx f(y_0)$ .

**transgalactic** · August 2nd 2011, 08:36 AM

ahhh i forgot about continuettyy
thanks

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