# 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

##### Topic

Calculus

##### Difficulty Level

6 out of 10

##### Suggested Book

Calculus in one variable by I.A.Maron

# Start with hints

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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