How to find the limit of $\frac{(1-x)(1-2x)(1-3x)...(1-85x)-1}{x}$? I am struggling to find out how to correctly write the steps to solve the limit of this series:
$$\lim_{x\to 0}=\frac{(1-x)(1-2x)(1-3x)...(1-85x)-1}{x}$$
I know, that the result is -3655 and I also know why: Since the limit approaches zero, powers higher than $x^1$ will be zero, and elements without $x$ will cancel out $(1-1=0)$. Therefore, I only care about elements that have $x^1$, which can be simplified to
$$-x(1+2+3+4+...+85)$$
and that gives us the formula $n(n+1)/2$, hence the result. The $x$ will then cancel out with the one in the denominator, leaving me with $-3655$.
What I do not know is how to solve this problem properly, that is using proper steps and formation.
Can anyone provide any solutions or tips?
FIRST METHOD :
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ defined for all $x \in \mathbb{R}$ by
$$f(x)=(1-x)(1-2x)(1-3x)...(1-85x)$$
What you want to calculate is
$$\lim_{x \rightarrow 0} \frac{f(x)-1}{x} = \lim_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$$
By definition of the derivative, you deduce $$\lim_{x \rightarrow 0} \frac{f(x)-1}{x} = f'(0)$$
which can be calculated, if you are courageous.
SECOND METHOD :
The function $f$ is polynomial, and its constant coefficient is $1$. That means that you can write $f(x)=1+\sum_{k=1}^{85} a_k x^k$. Then you see that
$$\frac{f(x)-1}{x}  = \sum_{k=1}^{85} a_k x^{k-1}$$
so the limit when $x$ tends to $0$ is equal exactly to $a_1$, i.e. the coefficient of $x$ in the polynomial function $f$. It is easy to see that this coefficient is equal to
$$a_1=- \sum_{k=1}^{85} k = -3655$$
With the same $ f(x) $ defined above and
$$g(x)=\ln(f(x))\;for \; x&lt;\frac{1}{85}$$
we have
$$f'(x)=f(x)g'(x)$$
with
$$g'(x)=\frac{-1}{1-x}-\frac{2}{1-2x}-...-\frac{85}{1-85x}$$
thus
$$f'(0)=f(0)g'(0)=-1-2-...-85=-\frac{85\cdot86}{2}$$