User:EGM5241.S11.team5.cavalcanti/HW1

Problem 1 - L'Hôpital's Rule
Given: $$ f(x)=\frac{e^{x}-1}{x}\ $$. Find: 1. $$ lim_{x\rightarrow 0} f(x) $$. 2. Plot

Solution
According to L'Hôpital's Rule, if we have and the denominator is equal to 0, we evaluate the limit by  Therefore, $$\frac{\mathrm{d} (e^{x}-1)}{\mathrm{d} x}=e^{x}$$, and $$ \frac{\mathrm{d} (x)}{\mathrm{d} x}=1$$. By applying L'Hôpital's Rule, we get: $$ \lim_{x\rightarrow 0}\frac{e^{x}-1}{x}=\lim_{x\rightarrow 0}\frac{e^{x}}{1}=\frac{1}{1}=1 $$

Plot
The plot is shown on the domain [0,1]:

Octave Code
octave-3.2.4.exe:1> x=0:0.1:1; octave-3.2.4.exe:2> y=(exp(x)-1)/x; octave-3.2.4.exe:3> plot(x,y,"@13")

Problem 2: Taylor Series
Given: $$ f(x)=\frac{e^{x}-1}{x}$$. Find: $$ p(x)$$ and $$ R_{n+1}(x)$$.

$$p(x)$$
To find $$p(x)$$, we need to take the Taylor Series, as seen in Meeting 3. We start out by taking the Taylor Series of $$e^{x}$$, as shown: $$x_0=0$$ and $$ \frac{d e^{x}}{dx}=e^{x}$$. Therefore, the Taylor Series Expansion is: $$p(x)= 1+\frac{x}{1!}+\frac{x^{2}}{2!}+ ... +\frac{x^{n}}{n!}$$. Subtracting 1, we then get: $$p(x)= \frac{x}{1!}+\frac{x^{2}}{2!}+ ... +\frac{x^{n}}{n!}$$. Finally, dividing by x, we get the final answer: $$p(x)=1+\frac{x}{2!}+\frac{x^{2}}{3!}+ ... +\frac{x^{n-1}}{n!}$$