User:Egm6341.s11.team3.elango

Numerical Methods 1: Team 3
{| class="toccolours collapsible collapsed" style="width:100%" ! Submission 1: Jan 5 - Jan 14

Find
Evaluate $$\lim _{x\mapsto 0}f\left ( x \right )$$ and plot $$f\left ( x \right ), x\in \left [ 0,1 \right ]$$

Given

 * $$f\left ( x \right )=\frac{e^{x}-1}{x}$$

Solution
In its simplest form, l'Hôpital's rule states that for functions f and g:

If $$\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 $$ or $$\pm\infty $$ and $$\lim_{x\to c}f'(x)/g'(x)$$ exists,

then $$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$$

REF : http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Hence, we differentiate the numerator and denominator individually to apply the limit

Numerator:
 * $$\frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x}-1 \right )|_{x=0}=e^{x}|_{x=0}=1$$

Demoninator:
 * $$\frac{\mathrm{d} }{\mathrm{d} x}\left ( x \right )|_{x=0}=1$$

Therefore
 * {| style="width:100%" border="0" align="left"


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\lim _{x\mapsto 0}\frac{e^{x}-1}{x}=\frac{1}{1}=1 $$ $$
 * $$\displaystyle
 * }
 * }

HW 1.4

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{| class="toccolours collapsible collapsed" style="width:100%" ! Submission 2: Jan 5 - Jan XX

HW 2.4

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