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

Can someone please help me with this? I'm not able to make any progress. I'm not allowed to use l'Hopital's rule.

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  • $\begingroup$ Are you permitted to use Taylor's Theorem? $\endgroup$ – Mark Viola Sep 16 at 20:41
  • $\begingroup$ @MarkViola Yes. $\endgroup$ – Tapi Sep 16 at 20:48
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Sketch: Recall $\cos t \sim 1-t^2/2$ as $t\to 0.$ (L'Hopital gives this if you like.) So in our expression, note that $1/n^x\to 0$ for each fixed $x.$ So let's be lazy and write $\cos (1/n^x) = 1-(1/n^x)^2/2.$ It's not correct, but it's going to be close and we want to get a feel for what's happening here. Then

$$(\cos (1/n^x))^n = [(1-(1/(2n^x))^{2n^x}]^{n/(2n^x)}.$$

Inside the brackets we have the limit $1/e.$ Now if $x=1,$ the outside power is just $1/2.$ Seems like the limit should then be $[1/e]^{1/2}.$ What happens if $0<x<1?$ What happens if $x>1?$

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Note that $\ln{\cos(n^{-x})^n}=n(\ln \circ \cos)(n^{-x})$.

Let $g(z)=\ln{\cos(z^{1/2})}$ for $0\leq z<1$, $g$ is smooth negative outside zero, continuous and $g(0)=0$. Moreover, as $z \rightarrow 0$, $g(z) \sim \cos(z^{1/2})-1 \sim -2\left(\sin{z^{1/2}/2}\right)^2\sim -z/2$, so $g$ is differentiable at $0$ with $g’(0)=-1/2$.

Now, $f(x)$ is the limit of $e^{ng(n^{-2x})}$. But $ng(n^{-2x}) \sim -n^{1-2x}/2$, so the argument goes to $\-infty$ when $x <1/2$, to $-1/2$ when $x=1/2$, and to $0$ when $x > 1/2$. So $f$ is $0$ on $(0,1/2)$, $e^{-1/2}$ at $1/2$, $1$ on $(1/2,\infty)$.

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