Wasted Expansion

Relevant wiki: Derivative by First Principle

As pointed out by Sambhrant Sachan in the solution discussion, we can evaluate this limit by applying the derivatives by first principle . That is, $\displaystyle f'(x) = \lim_{h\to0}\dfrac{f(x+h) - f(x)}h$ .

Let $f(x) = x^5$ , then $f(x+h) = (x + h)^5$ and $f(x-h) = (x-h)^5$ . So, $f(1+ h) - f(1-h) = (1+h)^5 - (1-h)^5$ .

We have

$\begin{aligned} \lim_{h\to0} \dfrac{(1+h)^5 - (1-h)^5}h &=& \lim_{h\to0} \dfrac{f(1+ h) - f(1-h)}h \\ &=& \lim_{h\to0} \dfrac{(f(1+ h) - f(1)) -( f(1-h) - f(1))}h \\ &=& \lim_{h\to0} \dfrac{f(1+h) - f(1)}h - \lim_{h\to0} \dfrac{f(1-h) - f(1)}h \\ \end{aligned}$

For the second limit, we can let $h = -j$ , then $\displaystyle \lim_{h\to0} \dfrac{f(1-h) - f(1)}h = - \lim_{j\to0} \dfrac{f(1+j) - f(1)}j$ .

And we can finish where we left off,

$\begin{aligned} \lim_{h\to0} \dfrac{(1+h)^5 - (1-h)^5}h &=& \lim_{h\to0} \dfrac{f(1+h) - f(1)}h + \lim_{j\to0} \dfrac{f(1+j) - f(1)}j \\ &=& \lim_{h\to0} \dfrac{f(1+h) - f(1)}h + \lim_{h\to0} \dfrac{f(1+h) - f(1)}h \\ &=& 2 \lim_{h\to0} \dfrac{f(1+h) - f(1)}h \\ &=& 2 f'(1) = 2 \left( \left . \dfrac{d}{dx} x^5 \right |_{x=1} \right) = 2 \cdot 5 \cdot (1)^4 = \boxed{10} . \end{aligned}$

I hate the L'Hopital's Rule a bit. So, I applied the following:

$\displaystyle \lim _{ h\to0 }{ \frac { ({ 1+h) }^{ 5 }-{ (1-h) }^{ 5 } }{ h } }$

$\displaystyle = 2\times \lim _{ h\to0 }{ \frac { ({ 1+h) }^{ 5 }-{ (1-h) }^{ 5 } }{ 2h } }$

$\displaystyle = 2\times \lim _{ h\to0 }{ \frac { ({ 1+h) }^{ 5 }-{ (1-h) }^{ 5 } }{ 1+h-1+h } }$

$\displaystyle = 2\times \lim _{ h\to0 }{ \frac { ({ 1+h) }^{ 5 }-{ (1-h) }^{ 5 } }{ (1+h)-(1-h) } }$

Observe that $h \rightarrow 0 \implies 1+h \rightarrow 1$ and $-h \rightarrow 0 \implies 1-h \rightarrow 1$ . Hence, we have $1+h \rightarrow 1-h$ . Now our limit advances to:

$\displaystyle = 2\times \lim _{ (1+h)\to(1-h) }{ \frac { ({ 1+h) }^{ 5 }-{ (1-h) }^{ 5 } }{ (1+h)-(1-h) } }$

Recall that $\displaystyle \lim _{ x\to a }{ \frac { { x }^{ n }-{ a }^{ n } }{ x-a } } =n{ a }^{ n-1 }$ (You can easily prove it by bashing a little with the binomial theorem). Applying this law here with $x=1+h$ and $a=1-h$ , our limit evolves to:

$= 2 \times 5(1-h)^{4}$

Recall that $h$ was approaching $0$ . Hence substituting it in the limit our answer comes $\boxed{10}$ .

Since it's a 0/0 form:

We can use L'Hopital Rule... By differentiating the numerator and putting h=0, numerator =10 and denominator =1

HENCE ANSWER = 10

A different solution from First Principles is to factor the expression, using the fact that $x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$ $\begin{aligned} \lim_{h\to0} \dfrac{ (1+h)^{5} - (1-h)^{5} }h & = \lim_{h\to0} \dfrac{[(1+h)-(1-h)][(1+h)^4+(1+h)^3(1-h)+...+(1-h)^4]}h \\ & = \lim_{h\to0} \dfrac{2h[(1+h)^4+ ... ]}h \\ \text{(cancel monomial \$h\$ on top and bottom, \& then substitute zero for \$h\$)} \\ & = 2[1^4+1^3\times 1 +1^2\times{1^2}+1 \times{1^3}+ 1^4] \\ & = 2\times 5 = \boxed{10} \end{aligned}$

the best method to do this would be using binomial theorm as h<<<1 ( as h-> 0) it would be 1-5h