Question #5d4b5

Question

1711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-04T21:11:00

#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#.