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# Understand the problem

Let $\{f_n\}$ be a sequence of functions from $\Bbb R$ to $\Bbb R$. where $f_n=\frac{\sqrt{1+(nx)^2}}{n}$. Then which of the following statement is true:
1. $\{f_n\}$ and $\{f_n'\}$ converge uniformly on $\Bbb R$.
2. $\{f_n'\}$ converges uniformly on $\Bbb R$ but $\{f_n\}$ does not.
3. $\{f_n\}$ converges uniformly on $\Bbb R$ but $\{f_n'\}$ does not.
4. $\{f_n\}$ converges uniformly to a differentiable function on $\Bbb R$.

##### Source of the problem
TIFR 2019 GS Part A, Problem 15
Analysis
Moderate
##### Suggested Book
Real Analysis, Bartle and Sherbert

Do you really need a hint? Try it first!

Here we will check the sup norm condition. See the hints in question 12 of GS 2019 in cheenta portal and try this question again.
Observe that $f_n \to f(x)=|x|$ and $M_n=Sup_{\{x \in \Bbb R\}}|f_n(x)-f(x)|$$\leq$$Sup_{\{x \in \Bbb R\}}$$(\frac 1n+|x|-|x|)=\frac 1n$

then $M_n \to 0$ as $n \to \infty$. Can you rule out any of the condition?

Calculate $f_n'$ and draw the conclusion.

$f_n'=\frac 1n \times \frac{1}{2\sqrt{1+(nx)^2}}\times 2xn^2=\frac{xn}{\sqrt{1+(nx)^2}}=\frac{xn}{n\sqrt{\frac {1}{n^2}+x^2}} \to \frac {x}{|x|}$

#### If a sequence of function is uniform convergent then a continuous sequence of functions will converge to a continuous function.

Now the given sequence of function is continuous but the limit is not, hence this is not a uniform convergence. So, option c is correct.

# Connected Program at Cheenta

#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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