Question #5d4b5

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Answer 1 · 2017-09-04T21:11:00

$4 t^{3}$ $4t^3$

You're asking for the limit:

$lim_(xrarrt)(x^4-t^4)/(x-t)$

Factor the numerator as a difference of squares:

$=lim_(xrarrt)((x^2)^2-(t^2)^2)/(x-t)=lim_(xrarrt)((x^2+t^2)(x^2-t^2))/(x-t)$

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

$=lim_(xrarrt)((x^2+t^2)(x+t)(x-t))/(x-t)$

The $x-t$ terms cancel:

$=lim_(xrarrt)(x^2+t^2)(x+t)$

We can evaluate for $x=t$ now:

$=(t^2+t^2)(t+t)=2t^2(2t)=4t^3$

Answer 2 · 2017-09-04T21:14:04

$4t^3$

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

$f'(t)=lim_(xrarrt)(f(x)-f(t))/(x-t)$

Where $f(x)=x^4$ , this becomes:

$f'(t)=lim_(xrarrt)(x^4-t^4)/(x-t)$

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