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## Understand the problem

Let $f$ : $(0,1) \rightarrow \mathbb{R}$ be defined by  $f(x) = \lim_{n\to\infty} cos^n(\frac{1}{n^x})$. (a) Show that $f$ has exactly one point of discontinuity. (b) Evaluate $f$ at its point of discontinuity.

##### Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2019
Calculus

6 out of 10

##### Suggested Book

Do you really need a hint? Try it first!

Try to determine the Form of the Limit. Show that the limit is of the form $1^\infty$. Hence, try to find the Functional Form.

$f(x) = \lim_{n\to\infty} (1 + (cos(\frac{1}{n^x}) – 1))^n = e^{(\lim_{n\to\infty}(cos(\frac{1}{n^x}) – 1).n}$.

${\lim_{n\to\infty}(cos(\frac{1}{n^x}) – 1).n = -\frac{1}{2}\lim_{n\to\infty}\frac{(sin^2(\frac{1}{2n^x}))}{(\frac{1}{2n^x})^2}.n^{(1-2x)} }$

Prove that  $f(x)$ = $\left\{ \begin{array}{ll} 0 & 0 < x < \frac{1}{2} \\ \frac{1}{\sqrt{e}} & x = \frac{1}{2} \\ 1 & x > \frac{1}{2} \\ \end{array} \right.$

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