# Show uniform convergence of $f_n (x)=\sqrt {\sin \left(\frac {x}{n}\right)+\cos\left(\frac {x}{n}\right) }$

How can I prove uniform convergence of $$f_n (x) =\sqrt {\sin \left(\frac {x}{n}\right)+\cos\left(\frac {x}{n}\right)}$$ for $$x \in \left[0,\frac{\pi}{2}\right]$$

I have been able to show its pointwise limit is $$1$$.

• Have have you tried? How did you show the pointwise limit? – Viktor Glombik Jan 26 at 14:16
• @ViktorGlombik, for any fixed $x$, by the continuity of the square root, sine and cosine functions, as n approaches infinity, we have $\sqrt (1)=1$ – user397197 Jan 26 at 14:19

Hint: Define $$f(x)= \sqrt{\sin(x) + \cos(x)}$$ This function is continuous and thus uniformly continuous on $$[0; \frac{\pi}{2}]$$.

Added: We have that for every $$x\in [0; \frac{\pi}{2}]$$ that $$\left\vert \frac{x}{n} - \frac{x}{m} \right\vert = \left\vert \frac{1}{n} - \frac{1}{m} \right\vert \cdot \vert x \vert \leq \frac{\pi}{2} \left\vert \frac{1}{n} - \frac{1}{m} \right\vert$$ Furthermore, we have $$\vert f_n(x) - f_m(x) \vert = \vert f(\frac{x}{n}) - f(\frac{x}{m}) \vert.$$ Of course, we don't need the uniform continuity (as shown in the comments by p4sch), but I consider it usefull to think this way. It allows one to also tackle different problems (like for example the uniform convergence of $$g_n(x)=\sqrt{x^2+\frac{1}{n}}$$ to the absolute value on the whole real line).

• I'm not sure how this should help to show that $f_n$ congerves uniformly. I think of $f_n(x)=x^n$, which are uniformly continuous on $[0,1]$ as well while it doesn't converge uniformly on $[0,1]$. – Mundron Schmidt Jan 26 at 14:21
• I know this but how does how does uniform continuity and uniform convergence relate in this case. Is there any theoren I am missing? – user397197 Jan 26 at 14:24
• You have $f_n(x) = f(x/n)$. Since $|f(0) -f(x)|< \varepsilon$ for all $|x|< \delta$, you get for all $1/n < \delta$ and $x \in [0,\pi/2]$ that $|f_n(x)| < \varepsilon$. – p4sch Jan 26 at 14:30
• @MundronSchmidt Honestly I do not see how your example is related to what I wrote. I consider a single function which reproduces the family and is uniformly continuous, but in your example only every individual function is uniformly continuous. Maybe you could elaborate on your comment? – Severin Schraven Jan 26 at 14:51
• @stackuser I did not add the second part so that you could try on your own. But it seems it was not clear enough. I guess now you should be able to fill in the gaps. Or do you want me to do it as well? – Severin Schraven Jan 26 at 14:53

Hint: use the inequalities $$|\sin(t)| \leq |t|$$ and $$|\cos(t)-1|\leq \frac{t^2}{2}$$.

• I am yet to know how to apply the hint. please guide me – user397197 Jan 26 at 15:39