How to determine $\lim_{h \to 0}\frac{g(h+1)-g(1)}{h}$ It is given that $g(x) = x^{20}$
Determine $$\lim_{h \to 0} \frac{g(h+1)-g(1)}{h}$$
Can someone give me a hint please? I worked it out to be so far as:
$$\lim_{h \to 0} \frac{(1+h)^{20}-1}{h}$$
The exponent of power $20$ is quite problematic. Any hints?
Note: I have to do it using first principles.
 A: Hint: $x^n-1=(x-1)(1+x+x^2+\cdots +x^{n-1})$. Set $x=1+h$.
A: Hint $$(1  +h)^{20} - 1 = h\dot\, ((h+1)^{19} +  \ldots + 1)$$
A: $[(1+h)^{20}-1]\\=[1+\binom{20}{1}.h+\binom{20}{2}.h^2+\cdots+\binom{20}{19}.h^{19}+h^{20}]-1\\=\space \binom{20}{1}.h+\binom{20}{2}.h^2+\cdots+\binom{20}{19}.h^{19}+h^{20}\\=\space h[\binom{20}{1}+\binom{20}{2}.h^1+\cdots+\binom{20}{19}.h^{18}+h^{19}] $
So,
$\lim_{h \to 0 }{{(1+h)^{20}-1}\over h}\\=\lim_{h \to 0 }{{h[\binom{20}{1}+\binom{20}{2}.h^1+\cdots+\binom{20}{19}.h^{18}+h^{19}]}\over h}\\=\lim_{h\to0}[\binom{20}{1}+\binom{20}{2}.h^1+\cdots+\binom{20}{19}.h^{18}+h^{19}]\\=\binom{20}{1}\\=20.$
A: For $(x+h)^n$ you can use the binomial formula
$$(x+h)^n = \sum_{k=0}^n\dbinom{n}{k}h^kx^{n-k}$$
With that you alloways get
$$\frac{\partial}{\partial x}x^n =\lim_{h \to 0} \frac{(x+h)^n-x^n}{h}=\lim_{h \to 0} \sum_{k=1}^n\dbinom{n}{k}h^{k-1}x^{n-k} $$
Only the term with $k=1$ survives in the sum. All the others go to zero.
Thus:
$$\frac{\partial}{\partial x}x^n =\dbinom{n}{1}h^{1-1}x^{n-1}=nx^{n-1} $$
