Evaluate the derivative of a function at $x=1$ and find a given limit at $\infty$ 
QUESTION: Suppose $f : \mathbb{R} \to \mathbb{R}$ is a function given by $$f(x)=e^{(x^{10}−1)}+(x− 1)^2 \sin(\frac{1}{x-1})$$
$(a)$ Find $f'(1)$.
$(b)$ Evaluate $$\lim_{u\to\infty} [100u-u\sum_{k=1}^{100}f(1+\frac{k}u)]$$


MY ANSWER: For the first part I have tried it in two ways, and the second in one.. let me explain what I have done-


Method 1/1

$f(x)=e^{(x^{10}-1)}+(x-1)^2\sin(\frac{1}{x-1})$
Therefore, $$f'(x)=10x^9e^{(x^{10}-1)}+(x-1)^2\cos\bigg(\frac{1}{x-1}\bigg)\frac{(-1)}{(x-1)^2}+\sin\bigg(\frac{1}{x-1}\bigg)2(x-1)$$ Now when $x→1$, since $\sin\theta$ is an oscillating number between $[-1,1]$ therefore $\sin\big(\frac{1}{x-1}\big)2(x-1)$ becomes $0$. Hence, $$f'(1)=10-\cos(\infty)$$ This is a finite value oscillating between $[9,11]$ but how do I find that? So here I am stuck..


Method 1/2

From the fundamental theorem of calculus $$f'(x)=\frac{f(x)-f(a)}{x-a}$$ Here $a=1$ and substituting $x$ to $1+h$ where $h→0$ we get- $$\lim_{h→0}\frac{f(1+h)-f(1)}{1+h-1}$$
$$\lim_{h→0}\frac{e^{[(1+h)^{10}-1]}+(1+h-1)^2\sin\bigg(\frac{1}{1+h-1}\bigg)}{h}$$
Now the second part becomes $\frac{h^2\sin(\frac{1}h)}{h}$ which is $\frac{\sin(\frac{1}h)}{\frac{1}h}$ when $h→0$. Therefore that is clearly $0$. Coming to the first part we get- $$\frac{e^{\big((1+h)^{10}-1\big)}}{h}$$ Expanding $(1+h)^{10}$ using binomial expansion, we obtain $$(1+h)^{10}= \binom{10}{0}+\binom{10}{1}h+\binom{10}{2}h^2+....$$ From here I cannot take it. The $\binom{10}{0}$ gets cancelled by $1$, but what will I do with the rest? I did a little further although illegally $:$P, that since $h→0$ therefore I completely ignored the terms containing $h^2$ or higher powers of $h$. So we are left with $e^{10h}$. Expanding this using Taylor expansion (Maclaurin to be precise) we get $$e^{10h}= 1 + 10h+ \frac{(10h)^2}{2!} + \frac{(10h)^3}{3!} +......$$ Since we have an $h$ in the denominator, only $10h$ becomes $10$ and remaining all $0$ except $1$ which becomes $\frac{1}h$. As you can see, $\frac{1}h$ creates problem here.

we remember that we did illegal assumptions, so mathematics always arrests anyway $:$P

Any help in this part?

Coming to the second part of the question-

Method 2/1

$$\lim_{u→\infty}[100u-u\sum_{k=1}^{100}f\bigg(1+\frac{k}u\bigg)]$$
Let's break this and see it in parts..

second part

$$\bigg(1+\frac{k}u-1\bigg)^2\sin\bigg(\frac{1}{1+\frac{k}u-1}\bigg)$$ $$\implies \bigg(\frac{k}u\bigg)^2\sin\bigg(\frac{u}k\bigg)$$ Now when $u→\infty$ this becomes $0$ due to clear reasons stated previously.. So every term of this summation becomes $0$.

first part

$$e^{\big((1+\frac{k}u)^{10}-1\big)}$$
When $u→\infty$ this becomes $e^{(1-1)}$ or $e^0$ therefore $1$. Therefore the summation upto $100$ terms becomes nothing but $100$. Therefore, our limit becomes $$\lim_{u→\infty}[100u-100u]$$ $$=0$$ Am I correct? If not where is the mistake that I have committed?
Any help will be much appreciated. Thank you so much.
 A: Let $$g(x) = e^{x^{10}-1}$$ and $$h(x) = (x-1)^2 \sin (x-1)^{-1}.$$  Consider these separately.  We have no problems with $g'(1) = 10$ with the usual methods.  However, $h$ is problematic.  We note that in order to make $h$ continuous at $x = 1$, we must choose $$h(1) = \lim_{x \to 1} h(x) = 0.$$  Next, we write $$h'(1) = \lim_{x \to 1} \frac{h(x) - h(1)}{x - 1} = \frac{h(x)}{x-1} = \lim_{x \to 1} (x-1) \sin (x-1)^{-1} = 0.$$  So $f'(1) = g'(1) + h'(1) = 10$.  For the second part, as $u \to \infty$, $f(1 + k/u)$ for $k \in \{1, 2, \ldots, 100\}$ tends toward $f(1)$, so this suggests writing
$$\lim_{u \to \infty} \left( 100u - u \sum_{k=1}^{100} f\left(1 + \frac{k}{u}\right) \right) =  \sum_{k=1}^{100} \lim_{u \to \infty} u \left(1 - f\left(1 + \frac{k}{u}\right)\right).$$  So we consider $$k \lim_{v \to 0^+} \frac{1 - f(1 + kv)}{kv}.$$  From here, I leave the rest as an exercise, keeping in mind that $g$ should not pose any problems, and to use similar reasoning we employed for $h$ to evaluate the limit. 
A: For the first part we need to use $f(1)=1$ by continuity giving
\begin{align}
\lim_{h\to0}\frac{f(1+h)-f(1)}h
&=\lim_{h\to0}\frac{e^{10h+o(h)}+o(h)-1}h\\
&=\lim_{h\to0}\frac{10h+o(h)}h\\
&=\lim_{h\to0}(10+o(1))\\
&=10\\
\end{align}
So we have $f'(1)=10$. Then using Taylor's theorem the second part is given by
\begin{align}
\lim_{u\to\infty}\left(100u-u\sum_{k=1}^{100}f\left(1+\frac{k}u\right)\right)
&=\lim_{u\to\infty}\left(100u-u\sum_{k=1}^{100}\left(f(1)+f'(1)\cdot\frac{k}u+o\left(\frac1u\right)\right)\right)\\
&=\lim_{u\to\infty}\left(100u-u\left(100\cdot f(1)+\frac{100\cdot(100+1)}{2\cdot u}\cdot f'(1)+o\left(\frac1u\right)\right)\right)\\
&=\lim_{u\to\infty}(-50500+o(1))\\
&=-50500\\
\end{align}
