An ISI Math

Fisrt assume that $f$ is differentiable at $x=1$ and that $f'(1)=10$

Then, $\forall k \in \mathbb{N}, f\left(1+\dfrac{k}{x}\right) \underset{x\rightarrow +\infty}{=} 1 + 10\dfrac{k}{x} + o\left(\dfrac{1}{x}\right)$

So, because the sum is finite,

$\begin{aligned} x \displaystyle \sum_{k=1}^{100} f\left(1+\dfrac{k}{x}\right) & \underset{x\rightarrow +\infty}{=} x \displaystyle \sum_{k=1}^{100} \left[ 1 + 10\dfrac{k}{x} + o\left(\dfrac{1}{x}\right) \right] \\ & \underset{x\rightarrow +\infty}{=} 100x + \displaystyle \sum_{k=1}^{100} 10k \quad + o\left(1\right)\\ & \underset{x\rightarrow +\infty}{=} 100x + 50500 + o\left(1\right) \end{aligned}$

Thus the answer is $\boxed{-50500}$

Proof that $f$ is differentiable at $x=1$ and that $f'(1)=10$

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ defined by $g(x)=x^2\sin(\dfrac{1}{x})$ if $x \neq 0$ and $g(0)=0$ .

$g$ is continuous and differentiable at $x=1$ because:

$|g(x)| \leq x^2 \underset{x\rightarrow 0}{\longrightarrow} 0=g(0)$

And $| \dfrac{g(x)-g(0)}{x-0} | = |x\sin(\dfrac{1}{x})| \leq |x| \underset{x\rightarrow 0}{\longrightarrow} 0$

Therefore $g$ is differentiable at $x=1$ and $g'(0)=0$ . So obviously, $h=g\circ (id-1)$ is differentiable at $x=1$ and $h'(1)=0$ .

This proves that $f$ is differentiable at $x=1$ and $f'(1) = \dfrac{ d e^{x^{10}-1} }{dx} |_{x=1} = 10 x^9 e^{x^{10}-1}|_{x=1}=10 \quad \square$