Question #e3f75

2 Answers

The limit is #10x^4-15x^2#

Explanation:

Let's look at the numerator we have that

#2(x+h)^5-2x^5=2*[(x+h)^5-x^5]=2*[h^5+5 h^4 x+10 h^3 x^2+10 h^2 x^3+5 h x^4]=2*h*[h^4+5h^3x+10h^2x^2+10hx^3+5x^4]#

and

#5x^3-5(x+h)^3=5*[x^3-(x+h)^3]=5*[-h^3-3 h^2 x-3 h x^2 ]=5*h*[-h^2-3hx-3x^2]#

Hence the limit becomes

#lim_(h->0) [2*h*[h^4+5h^3x+10h^2x^2+10hx^3+5x^4]+5*h*[-h^2-3hx-3x^2]]/[h]= lim_(h->0) 2*[h^4+5h^3x+10h^2x^2+10hx^3+5x^4]+5*[-h^2-3hx-3x^2]= 10*x^4-15x^2#

Jul 7, 2016

#10x^4 - 15x^2#

Explanation:

So it looks like you're using the limit definition of the derivative to find #d/(dx) (2x^5 - 5x^3)#. We know this will be #10x^4 - 15x^2# so it's a nice sense check when we use limits.

We're going to need to expand the brackets, in this case we'll use the binomial theorem.

#(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k#

where #((n),(k)) = (n!)/(k!(n-k)!)#

#(x+h)^5 = sum_(k=0)^5 ((5),(k)) x^(5-k)y^k#

#= ((5),(0))x^5h^0 + ((5),(1))x^4h^1 + ((5),(2)) x^3h^2 + ((5),(3))x^2h^3 + ((5),(4))x^1h^4 + ((5),(5))x^0h^5#

# = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5#

Similarly;

#(x+h)^3 = ((3),(0))x^3h^0 + ((3),(1))x^2h + ((3),(2))xh^2 + ((3),(3))x^0h^3#

#(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3#

Collating all this into the initial function we get:

#lim_(h->0) 1/h*(2[x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5] - 5[x^3 + 3x^2h + 3xh^2 + h^3] -2x^5 + 5x^3)#

Some terms cancel out

#lim_(h->0) 1/h*(color(red)(2x^5)+10x^4h+20x^3h^2+20x^2h^3+10xh^4+2h^5 color(blue)(- 5x^3) -15x^2h - 15xh^2 - 5h^3color(red)( - 2x^5) + color(blue)(5x^3))#

Now multiplying through by #1/h#

#lim_(h->0) (10x^4+20x^3h+20x^2h^2+10xh^3+2h^4 -15x^2 - 15xh)#

As #h->0# any term with h in it will go to zero and would you look at that! It's only gone and left us with

#10x^4 - 15x^2#

Pleasingly, this is what we expected so we can be fairly confident there hasn't been any errors in the calculation.