Limits , Continuity ,Derivatives

Limits  

  The functions  $f(x)$ is said to have the $limit$ $l$ as $x$ approaches $a$, abbreviated $\displaystyle\lim_{x\to a} f(x)=l$, if given any number $\epsilon >0$ we can find a number $\delta >0$ such that $|f(x)-l| < \epsilon $ whenever $0<|x-a|<\delta$

     Note that $|p|,i.e.$ the $absolute value$ of $p,$ is equal to $p$ if $p>0,-p$ if $p<0$ and $0$ if $p=0$.

     Example $\displaystyle\lim_{x\to 1} (x^2-4x+8) =5, \displaystyle\lim_{x\to 2}\cfrac{x^2-4}{x-2}=4, \displaystyle\lim_{x\to o}\cfrac{\sin x}{x}=1$

     If $\displaystyle\lim_{x\to a}f_1(x)=l_1$ $\displaystyle\lim_{x\to a}f_2 (x)=l_2$ then we have the following theorems on limits .

$(a)$ $\displaystyle\lim_{x\to a}[f_1(x)\pm f_2(x)] =\displaystyle f_1(x)\pm \displaystyle\lim_{x\to a}f_2(x)=l_1\pm l_2$

$(b)$ $\displaystyle\lim_{x\to a}[f_1 (x)f_2(x)] =\left[\displaystyle\lim_{x_a}f_1(x) \right]\left[\displaystyle\lim_{x\to a}f_2 (x)\right] =l_1 l_2$

$(c)$ $\displaystyle\lim_{x\to a}\cfrac{f_1 (x)}{f_2 (x)}=\cfrac{\displaystyle\lim_{x\to a}f_1 (x)}{\displaystyle\lim_{x\to a}f_2 (x)}=\cfrac{l_1}{l_2}$ if $l_2\neq 0 $

Continuity 

     The function $f(x)$ is said to be $continuous$ at $a$ if $\displaystyle\lim_{x\to a} f(x)=f(a)$.

Example $f(x)=x^2-4x+8$ is continuous at $x=1$ However ,if  $ f(x)=\begin{cases} \cfrac{x^2-4}{x-2} & x\neq 2 \\ 6 & x=2   \end{cases}$ 

Then $f(x)$ is not continuous [or is discontinuous ] at $x=2$ and $x=2$ is called $a$ discontinuous of $f(x)$ 

     If $f(x)$ is continuous at each point of an interval such as $x_1\leqslant\leqq x\leqslant\leqq x_2$ or $x_1< x\leqslant\leqq x_2, x$ etc. it is said to be continuous in the interval .

     If $f_1(x)$ and $f_2(x)$ are continuous in an interval then $f_1(x)\pm f_2(x),f_1(x)f_2(x)$ and $f_1(x)/f_2(x)$ where $f_2(x)\neq 0$ are also continuous in the interval.

Derivatives

     The $derivatives$ of $y=f(x)$ at a point $x$ is defined as

$f'(x)=\displaystyle\lim_{h\to 0}\cfrac{f(x+h)-f(x)}{h}=\displaystyle\lim_(\bigtriangleup x\to 0)\cfrac{\bigtriangleup y}{\bigtriangleup x}=\cfrac{\mathrm{d}y}{\mathrm{d}x}$

Where $h=\bigtriangleup x, \bigtriangleup y=f(x+h)-f(x)=f(x+\bigtriangleup x)-f(x)$ provided the limit exists .

The $differential$ of $y=f(x)$ is defined by

$\mathrm{d}y=f'(x)\mathrm{d}x$ where $\mathrm{d}x=\bigtriangleup x$      

The process of finding derivatives is called $differentiation$ . By taking derivatives of $y'=\mathrm{d}y/\mathrm{d}x=f'(x)$ we can find second. third  and higher order derivatives, denoted by $y'' =\mathrm{d}^2 y/\mathrm{d}x^2 =f''(x),y'''=\mathrm{d}^3y/\mathrm{d}x^3=f'''(x)$ etc.

     Geometrically the derivative of a function $f(x)$ at a point represents the $slope $ of the  $tangent line$ drawn to the curve $y=f(x)$ at the point .

     If a function has a derivative at a point, then it is continuous at the point. However, the converse is not necessarily .

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