# Find? lim xrarr c ((x^3 - c^3)/ (x-c))

Jun 17, 2018

$3 {c}^{2}$

#### Explanation:

${\lim}_{x \to c} \left(\frac{{x}^{3} - {c}^{3}}{x - c}\right)$

First, factor ${x}^{3} - {c}^{3}$
${x}^{3} - {c}^{3} = \left(x - c\right) \left({x}^{2} + x c + {c}^{2}\right)$

${\lim}_{x \to c} \left(\frac{\left(x - c\right) \left({x}^{2} + x c + {c}^{2}\right)}{x - c}\right)$
$\iff {\lim}_{x \to c} \left({x}^{2} + x c + {c}^{2}\right)$
$\iff {c}^{2} + c \cdot c + {c}^{2}$
$\iff 3 {c}^{2}$

Jun 18, 2018

$3 {c}^{2}$.

#### Explanation:

If one is familiar with the following Standard Form of Limit :

${\lim}_{x \to a} \frac{{x}^{n} - {a}^{n}}{x - a} = n {a}^{n - 1}$,

then, the required limit $3 {c}^{2}$ follows immediately.

Here is another way to get the limit :

Let, $x = c + h , \text{ so that, (x-c)=h, and, as } x \to c , h \to 0$.

Further, $\frac{{x}^{3} - {c}^{3}}{x - c} = \frac{{\left(c + h\right)}^{3} - {c}^{3}}{h}$,

$= \frac{\cancel{{c}^{3}} + {h}^{3} + 3 c h \left(c + h\right) \cancel{- {c}^{3}}}{h}$,

$= \frac{\cancel{h} \left\{{h}^{2} + 3 c \left(c + h\right)\right\}}{\cancel{h}}$,

$\therefore \text{The Reqd. Lim.} = {\lim}_{h \to 0} \left\{{h}^{2} + 3 c \left(c + h\right)\right\}$,

$= {0}^{2} + 3 c \left(c + 0\right)$,

$= 3 {c}^{2}$, as Martin C. has readily derived!