there is $\displaystyle lim_{x->x_{0}}g(x)=y_{0}$, $\displaystyle f(x)$is continues in $\displaystyle y_{0}$

point. i need to prove that $\displaystyle lim_{x->x_{0}}f\circ g(x)=f(lim_{x->x_{0}}g(x))=f(y_{0})$

by epsilon delta definition.

i need to follow __only__ my books prove.help me understand it.

the books proof:

in order to prove $\displaystyle lim_{x->x_{0}}f\circ g(x)=f(lim_{x->x_{0}}g(x))=f(y_{0})$

we will use the definition that for $\displaystyle N_{\varepsilon}^{*}(f(y_{0}))$

surrounding we have $\displaystyle N_{\delta}^{*}(x)$ surrounding for which every

$\displaystyle x\in N_{\delta}^{*}(x)$ $\displaystyle f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$.

**first step:**

$\displaystyle f(x)$is continues in $\displaystyle y_{0}$ point. so $\displaystyle lim_{x->x_{0}}f(x)=f(y_{0})$

and for $\displaystyle N_{\varepsilon}^{*}(f(y_{0}))$ surrounding we have $\displaystyle N_{\delta}^{*}(x)$

surrounding for which every

$\displaystyle x\in N_{\delta}^{*}(x)$ $\displaystyle f(x)\in N_{\varepsilon}^{*}(f(y_{0}))$.

__second step:__

it is given that $\displaystyle lim_{x->x_{0}}g(x)=y_{0}$ then for $\displaystyle x\in N_{\delta_{1}}^{*}(x)$

$\displaystyle g(x)\in N_{\varepsilon}^{*}(y_{0})$

so $\displaystyle f((g(x)))$$\displaystyle \in N_{\varepsilon}^{*}(f(y_{0}))$.

cant understand this last step why $\displaystyle f((g(x)))$$\displaystyle \in N_{\varepsilon}^{*}(f(y_{0}))$

is the resolt of the last two steps??