# Finding $\lim_{h \to 0} \frac{(x+h)^4 -x^4}{h}$ [closed]

How would i find the limit of h as it approaches 0 for an expression like below

$$\lim_{h \to 0} \frac{(x+h)^4 -x^4}{h}$$

• Have you learnt en.wikipedia.org/wiki/Binomial_theorem Nov 11, 2017 at 4:52
• $$a^4-b^4=(a-b)(a+b)(a^2+b^2)$$ Nov 11, 2017 at 5:29
• What you need is a simple binomial expansion. Nov 11, 2017 at 5:31

By definition $$\lim_{h \to 0} \frac{(x+h)^4 -x^4}{h}=(x^4)'=4x^3.$$ Also, $$\lim_{h \to 0} \frac{(x+h)^4 -x^4}{h}= \lim_{h \to 0} \frac{x^4+4x^3h+6x^2h^2+4xh^3+h^4 -x^4}{h}=4x^3$$

Let $a = (x+h)$ and $b=x$ then $h = a-b$ and $h \to 0$ as $a \to b$

So we have $\displaystyle \lim_{a \to b} \frac{a^4-b^4}{a-b} = \lim_{a \to b} [(a+b)(a^2+b^2)] = 4b^3 = 4x^3.$

note that $$(x+h)^4-x^4={h}^{4}+4\,{h}^{3}x+6\,{h}^{2}{x}^{2}+4\,h{x}^{3}$$

I once heard someone claiming everything can be proved by induction, so here it is, let's prove something more general and make the problem be a particular instance of the claim for $n=4$.

$\mathcal P(n):\lim\limits_{h\to 0}\dfrac{(x+h)^n-x^n}h=nx^{n-1}$

$\mathcal P(1): \dfrac{(x+h)-x}h=\dfrac hh=1\to 1=1\times x^0\quad\color{green}\checkmark$

Now assuming $\mathcal P(n)$ we get:

$\dfrac{(x+h)^{n+1}-x^{n+1}}h=\dfrac{((x+h)^n-x^n)(x+h)+hx^n}h=\underbrace{\left(\dfrac{(x+h)^n-x^n}h\right)}_{\to nx^{n-1}}\underbrace{(x+h)}_{\to x}+x^n\to (nx^{n-1})x+x^n=(n+1)x^n\quad\color{green}\checkmark$

So $\mathcal P(n)$ is true for all $n>0$.

In particular for $n=4$, the limit is $4x^3$.