Try this beautiful Subjective Calculus Problem appeared in ISI Entrance - 2019.
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.
Calculus
Limit , Continuity
Suggested Reading: IIT mathematics by Asit Das Gupta
Source of the Problem: ISI UG Entrance - 2019 , Subjective problem number - 2
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:
\[f(x)=\lim _{n \rightarrow \infty} \cos ^{n}\left(\frac{1}{n^{x}}\right)\]
\[=e^{\lim _{n \rightarrow \infty} n \log \cos \left(\frac{1}{n^{x}}\right)}\]
\[=e^{\lim _{n \rightarrow \infty}\frac{ n \log [1-2 \sin ^{2}(\frac{1}{2 n^{x}})]}{-2 \sin ^{2}(\frac{1}{2 n^{x}})}(-2 \sin ^{2}(\frac{1}{2 n^{x}}))}\]
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 \(x= \frac{1}{2}\)
ISI Entrance Program at Cheenta
Try this beautiful Subjective Calculus Problem appeared in ISI Entrance - 2019.
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.
Calculus
Limit , Continuity
Suggested Reading: IIT mathematics by Asit Das Gupta
Source of the Problem: ISI UG Entrance - 2019 , Subjective problem number - 2
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:
\[f(x)=\lim _{n \rightarrow \infty} \cos ^{n}\left(\frac{1}{n^{x}}\right)\]
\[=e^{\lim _{n \rightarrow \infty} n \log \cos \left(\frac{1}{n^{x}}\right)}\]
\[=e^{\lim _{n \rightarrow \infty}\frac{ n \log [1-2 \sin ^{2}(\frac{1}{2 n^{x}})]}{-2 \sin ^{2}(\frac{1}{2 n^{x}})}(-2 \sin ^{2}(\frac{1}{2 n^{x}}))}\]
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 \(x= \frac{1}{2}\)
ISI Entrance Program at Cheenta
