# Question #5d4b5

Sep 4, 2017

$4 {t}^{3}$

#### Explanation:

${\lim}_{x \rightarrow t} \frac{{x}^{4} - {t}^{4}}{x - t}$

Factor the numerator as a difference of squares:

$= {\lim}_{x \rightarrow t} \frac{{\left({x}^{2}\right)}^{2} - {\left({t}^{2}\right)}^{2}}{x - t} = {\lim}_{x \rightarrow t} \frac{\left({x}^{2} + {t}^{2}\right) \left({x}^{2} - {t}^{2}\right)}{x - t}$

Factor ${x}^{2} - {t}^{2}$ with the same method:

$= {\lim}_{x \rightarrow t} \frac{\left({x}^{2} + {t}^{2}\right) \left(x + t\right) \left(x - t\right)}{x - t}$

The $x - t$ terms cancel:

$= {\lim}_{x \rightarrow t} \left({x}^{2} + {t}^{2}\right) \left(x + t\right)$

We can evaluate for $x = t$ now:

$= \left({t}^{2} + {t}^{2}\right) \left(t + t\right) = 2 {t}^{2} \left(2 t\right) = 4 {t}^{3}$

Sep 4, 2017

$4 {t}^{3}$

#### Explanation:

Note the limit definition of the derivative for a function $f \left(x\right)$ at a point:

$f ' \left(t\right) = {\lim}_{x \rightarrow t} \frac{f \left(x\right) - f \left(t\right)}{x - t}$

Where $f \left(x\right) = {x}^{4}$, this becomes:

$f ' \left(t\right) = {\lim}_{x \rightarrow t} \frac{{x}^{4} - {t}^{4}}{x - t}$

Which fits what we're looking for! Thus, we only need to find $f ' \left(t\right)$. Well, $f ' \left(x\right) = 4 {x}^{3}$, so $f ' \left(t\right) = 4 {t}^{3}$.