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ISI 2019 Subjective Problem 2 | Removable Discontinuity

Try this beautiful Subjective Calculus Problem appeared in ISI Entrance - 2019.

Problem

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

Key Concepts

Calculus

Limit , Continuity

Suggested Book | Source | Answer

Suggested Reading: IIT mathematics by Asit Das Gupta

Source of the Problem: ISI UG Entrance - 2019 , Subjective problem number - 2

Try with Hints

Hint 1:

Observe that,

we have $$\frac{1}{n^x} \rightarrow 0$$

as x∈(0,) and n→∞.

So the indeterminate form of the given limit is $$1^{\infty}.$$

Hint 2:

$f(x)=\lim _{n \rightarrow \infty} \cos ^{n}\left(\frac{1}{n^{x}}\right)$

$=e^{\lim _{n \rightarrow \infty} n \log \cos \left(\frac{1}{n^{x}}\right)}$

=elimn→∞nlog[1−2sin2(12nx)]

$=e^{\lim _{n \rightarrow \infty}\frac{ n \log [1-2 \sin ^{2}(\frac{1}{2 n^{x}})]}{-2 \sin ^{2}(\frac{1}{2 n^{x}})}(-2 \sin ^{2}(\frac{1}{2 n^{x}}))}$

Hint 3:

As we have $$\frac{1}{n^x} \rightarrow 0$$ ,

$lim_{n\rightarrow \infty}-2 \sin ^{2}\left(\frac{1}{2 n^{x}}\right)=0.$

And we have the standard result :

$lim_{x \rightarrow 0}\frac{log(1+x)}{x} = 1$

Hint 4:

Therefore ,

$f(x)=e^{\lim _{n \rightarrow \infty} n [-2 \sin ^{2}(\frac{1}{2 n^{x}})]}$

$f(x)=e^{\lim _{n \rightarrow \infty} -2n [\frac{ \sin ^{2}(\frac{1}{2 n^x})}{(\frac{1}{2 n^{x}})^2}](\frac{1}{2 n^{x}})^2}$

And here again,

$$lim_{n \rightarrow \infty}\frac{1}{2n^x} \rightarrow 0 .$$

So $$f(x)$$ reduces to ,

$f(x) = e^{\lim _{n \rightarrow \infty}-2n (\frac{1}{2 n^{x}})^2}$

$f(x)=e^{-\frac{1}{2} \lim _{n \rightarrow \infty} n^{1-2 x}}$

Hint 5:

Therefore ,

$f(x) = \Bigg{\{}{\begin{matrix} 0 & , & x < \frac{1}{2}\\ e^{-\frac{1}{2}} & ,& x= \frac{1}{2} \\1 & , & x> \frac{1}{2} \end{matrix}}$

Discontinuity at $$x= \frac{1}{2}$$

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