# 经济和金融模型的基础|EMET1001 Foundations of Economic and Financial Models代写

1. Consider the function
$$f(x)=\left{\begin{array}{l} \frac{1}{x-1} \text { if } x \neq 1 \ 0 \text { if } x=1 \end{array}\right.$$(a) Does $\lim {x \rightarrow 5} f(x)$ exist? If $s 0$, what is it? Try and establish the validity of your answer formally using an epsilon-delta argunent. If it exists, does it equal $f(5)$ ? Is this function continuous at the point $x=5$ ? (b) Does $\lim {x \rightarrow 1} f(x)$ exist? If so, what is it? Try and establish the validity of your answer formally using an epsilon-delta arguneut. If it exists, does it equal $f(1)$ ? Is this function continuous at the point $x=1$ ?
(c) Is this function contimuous?

Solution:

Consider the function
$$f(x)= \begin{cases}\frac{1}{x-1} & \text { if } x \neq 1 \ 0 & \text { if } x=1\end{cases}$$
Part (a)
First of all, note that
$$\lim {x \rightarrow 5} f(x)=\lim {x \rightarrow 5}\left(\frac{1}{x-1}\right)=\frac{\lim {x \rightarrow \pi} 1}{\lim {x \rightarrow 5}(x-1)}=\frac{1}{4}=f(5) .$$
Thus we know that (i) $\lim {x \rightarrow 5} f(x)=\frac{1}{4}$, and (ii), the function is continuous at the point $x=5$. However, we also want to show that $\lim {x \rightarrow 5} f(x)=\frac{1}{4}$ formally, using an epsilon-delta proof. Note that
$$d(5, x)=\sqrt{(x-5)^{2}}=|x-5|,$$

and
$$d(f(5), f(x))=\sqrt{(f(x)-f(5))^{2}}=|f(x)-f(5)|=\left|f(x)-\frac{1}{4}\right| .$$
Suppose that we want $d(f(5), f(x))<\varepsilon$, where $\varepsilon>0$. Any problems will be caused by small values of $\varepsilon>0$. (Try and explain why this is the case.) Thus we can focus on the cases where
$$d(f(5), f(x))=\left|f(x)-\frac{1}{4}\right|=\left|\frac{1}{(x-1)}-\frac{1}{4}\right|$$
Note that
$$d(f(5), f(x))<\varepsilon \Longleftrightarrow\left|\frac{1}{(x-1)}-\frac{1}{4}\right|<\varepsilon_{.}$$ This requires that both $\frac{1}{(x-1)}-\frac{1}{4}<\varepsilon$ and $-\left(\frac{1}{(x-1)}-\frac{1}{4}\right)\frac{4}{1+4 \epsilon} \Longleftrightarrow x>\frac{4}{(1+4 c)}+1 \Longleftrightarrow x>1+\frac{4}{(1+4 \epsilon)} $$Note that$$ \left(\frac{1}{(x-1)}-\frac{1}{4}\right)<\varepsilon \Longleftrightarrow \frac{1}{(x-1)}-\frac{1}{4}>-\varepsilon $$We have$$ \frac{1}{(x-1)}-\frac{1}{4}>-\varepsilon \Longleftrightarrow \frac{1}{(x-1)}>\frac{1}{4}-\varepsilon \Longleftrightarrow \frac{1}{(x-1)}>\frac{1-4 \varepsilon}{4} . $$This requires that$$ x-1<\frac{4}{1-4 \epsilon} \Longleftrightarrow x<\frac{4}{(1-4 \epsilon)}+1 \Longleftrightarrow x<1+\frac{4}{(1-4 \epsilon)} $$Thus we know that$$ d(f(5), f(x))<\varepsilon \Longleftrightarrow 1+\frac{4}{(1+4 c)}<x<1+\frac{4}{(1-4 \epsilon)} $$Suppose that we waut d(5, x)<\delta. This reyuires that$$ d(5, x)<\delta \Longleftrightarrow|x-5|<\delta \Longleftrightarrow-\delta<x-5<\delta \Longleftrightarrow 5-\delta<x<5+\delta$\$

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