Limit to Function - ISI UG 2019 Subj Problem 2

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]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.

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I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2019
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Calculus

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]6 out of 10

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Calculus in one variable by I.A.Maron [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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Do you really need a hint? Try it first!

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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.

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

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

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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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Watch the video ( Coming Soon ... )

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Similar Problems

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