User:Egm6341.s11.team5.shin/HW

Problem statement
Let $$\displaystyle f(x)= \frac{e^x-1}{x}$$. Find $$\displaystyle \lim_{x \to 0}f(x)$$ and plot $$\displaystyle f(x)$$ for $$\displaystyle x \in [0,1] $$.

Solution
Using L'Hospital's Rule, the limit can be evaluated. L'Hospital's Rule:
 * $$\lim_{x \to 0}\frac{g(x)}{h(x)}=\lim_{x \to 0}\frac{g^{'}(x)}{h^{'}(x)}$$

Applying the Rule:

As a result, the limit is shown below.
 * $$\displaystyle g(x)=e^x-1$$ || $$\displaystyle g^{'}(x)=e^x$$
 * $$\displaystyle h(x)=x$$ || $$\displaystyle h^{'}(x)=1$$
 * }
 * $$\displaystyle h(x)=x$$ || $$\displaystyle h^{'}(x)=1$$
 * }
 * $$\lim_{x \to 0}\frac{e^x-1}{x}=\lim_{x \to 0}\frac{e^x}{1}$$
 * $$\lim_{x \to 0}\frac{e^x}{1}=\frac{1}{1}=1$$

Plot: Matlab code:

Problem statement
Find $$\displaystyle P_n(x)$$ and $$\displaystyle R_{n+1}(x)$$ of $$\displaystyle e^x$$ and $$\displaystyle \frac{e^x-1}{x}$$.

Solution
From the class note, $$\displaystyle P_n(x)$$ and $$\displaystyle R_{n+1}(x)$$ is defined as