 Try this problem from ISI B.Stat, B.Math Subjective Entrance Exam, 2017 Problem no. 3 based on Differentiability at origin.

Problem: Differentiability at origin

Suppose $f : \mathbb{R} \to \mathbb{R}$ is a function given by $$f(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 1 & \text{if x=1}\\ e^{(x^{10} -1)} + (x-1)^2 \sin \left (\frac {1}{x-1} \right ) & \text{if} x \neq 1\ \end{array}\right.$$

• Find f'(1)
• Evaluate $\displaystyle{\lim_{u \to \infty } \left [ 100 u – u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] }$

Discussion:

a)First of all we need to check whether $f'(1)$ exists or not.

We will proceed with the first principle.

Let us check the Right hand derivative(RHD) and Left hand derivative(LHD) of $f$ at $x=1$.

RHD at $x=1$ is

$\lim_{h\to0}\frac{f(1+h)-f(1)}{h}\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}+h^2\sin( \frac{1}{h})-1}{h}\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}-1}{h}+\lim_{h\to0}h\sin( \frac{1}{h})\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}-1}{(1+h)^{10}-1}\frac{(1+h)^{10}-1}{h}+0=10$

LHD at $x=1$ is

$\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}+h^2\sin( \frac{1}{-h})-1}{-h}\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}-1}{-h}+\lim_{h\to0}(-h)\sin( \frac{1}{-h})\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}-1}{(1-h)^{10}-1}\frac{(1-h)^{10}-1}{-h}+0=10$

Thus,LHD=RHD.

Hence $f'(1)$ exists and it is equal to $10$.

(b)

$\displaystyle{\lim_{u \to \infty } \left [ 100 u – u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] }$

As u becomes infinitely large k/u becomes arbitrarily small for finite value of k (clearly k is finite as we are interested in k=1 to 100).

Hence $f \left ( 1 + \frac {k}{u} \right )$ is nothing but f of (1 plus an infinitesimal positive quantity). This tells us $f \left ( 1 + \frac {k}{u} \right )$ is almost waiting to become the derivative of f at x=1. And we already know that such a derivative exists from part (a).

With this motivation, divide and multiply by $\frac{k}{u}$.

$\displaystyle{\lim_{u \to \infty } \left [ 100 u – u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] \\ =\lim_ {u \to \infty} \left [ 100 u -\sum_{k=1}^{100} k \frac{f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \right ]\\=\lim_ {u \to \infty} \left [ \sum_{k=1}^{100} k \frac{1-f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \right ]\\ =\sum_{k=1}^{100} k \lim_ {u \to \infty}\frac{f(1)-f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \\ =\sum_{k=1}^{100} k \times(- f'(1)) \\ = -10\times \left(\frac{100\times101}{2} \right )\\ =-50500 }$

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