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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

Calculus in one variable by I.A.Maron

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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