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.

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