# Evaluating $f:(0,\infty)\to\mathbb{R}$ defined by $f(x)=\lim_{n\to\infty}\cos^n\bigg(\frac{1}{n^x}\bigg)$ at the point of discontinuity [closed]

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.

• Are you permitted to use Taylor's Theorem? – Mark Viola Sep 16 at 20:41
• @MarkViola Yes. – Tapi Sep 16 at 20:48

## 2 Answers

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 $$0 What happens if $$x>1?$$

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 $$(1/2,\infty)$$.