# Question e3f75

The limit is $10 {x}^{4} - 15 {x}^{2}$

#### Explanation:

Let's look at the numerator we have that

$2 {\left(x + h\right)}^{5} - 2 {x}^{5} = 2 \cdot \left[{\left(x + h\right)}^{5} - {x}^{5}\right] = 2 \cdot \left[{h}^{5} + 5 {h}^{4} x + 10 {h}^{3} {x}^{2} + 10 {h}^{2} {x}^{3} + 5 h {x}^{4}\right] = 2 \cdot h \cdot \left[{h}^{4} + 5 {h}^{3} x + 10 {h}^{2} {x}^{2} + 10 h {x}^{3} + 5 {x}^{4}\right]$

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

$10 {x}^{4} - 15 {x}^{2}$

#### Explanation:

So it looks like you're using the limit definition of the derivative to find $\frac{d}{\mathrm{dx}} \left(2 {x}^{5} - 5 {x}^{3}\right)$. We know this will be $10 {x}^{4} - 15 {x}^{2}$ 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.

${\left(x + y\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {x}^{n - k} {y}^{k}$

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

${\left(x + h\right)}^{5} = {\sum}_{k = 0}^{5} \left(\begin{matrix}5 \\ k\end{matrix}\right) {x}^{5 - k} {y}^{k}$

= ((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#

$= {x}^{5} + 5 {x}^{4} h + 10 {x}^{3} {h}^{2} + 10 {x}^{2} {h}^{3} + 5 x {h}^{4} + {h}^{5}$

Similarly;

${\left(x + h\right)}^{3} = \left(\begin{matrix}3 \\ 0\end{matrix}\right) {x}^{3} {h}^{0} + \left(\begin{matrix}3 \\ 1\end{matrix}\right) {x}^{2} h + \left(\begin{matrix}3 \\ 2\end{matrix}\right) x {h}^{2} + \left(\begin{matrix}3 \\ 3\end{matrix}\right) {x}^{0} {h}^{3}$

${\left(x + h\right)}^{3} = {x}^{3} + 3 {x}^{2} h + 3 x {h}^{2} + {h}^{3}$

Collating all this into the initial function we get:

${\lim}_{h \to 0} \frac{1}{h} \cdot \left(2 \left[{x}^{5} + 5 {x}^{4} h + 10 {x}^{3} {h}^{2} + 10 {x}^{2} {h}^{3} + 5 x {h}^{4} + {h}^{5}\right] - 5 \left[{x}^{3} + 3 {x}^{2} h + 3 x {h}^{2} + {h}^{3}\right] - 2 {x}^{5} + 5 {x}^{3}\right)$

Some terms cancel out

${\lim}_{h \to 0} \frac{1}{h} \cdot \left(\textcolor{red}{2 {x}^{5}} + 10 {x}^{4} h + 20 {x}^{3} {h}^{2} + 20 {x}^{2} {h}^{3} + 10 x {h}^{4} + 2 {h}^{5} \textcolor{b l u e}{- 5 {x}^{3}} - 15 {x}^{2} h - 15 x {h}^{2} - 5 {h}^{3} \textcolor{red}{- 2 {x}^{5}} + \textcolor{b l u e}{5 {x}^{3}}\right)$

Now multiplying through by $\frac{1}{h}$

${\lim}_{h \to 0} \left(10 {x}^{4} + 20 {x}^{3} h + 20 {x}^{2} {h}^{2} + 10 x {h}^{3} + 2 {h}^{4} - 15 {x}^{2} - 15 x h\right)$

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

$10 {x}^{4} - 15 {x}^{2}$

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