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
Topic
Analysis
Difficulty Level
Moderate
Suggested Book
Real Analysis, Bartle and Sherbert

Start with hints

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)|\leqSup_{\{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|}

Can you recall any of the property of uniform convergence of sequence of functions which will call the conclusion?

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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