Originally Posted by

**Laurent** Ok, I assumed you let the function equal 1 at 0 since this is the limit (remember $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$), but you don't have to.

Suppose the convergence were uniform on $\displaystyle (0,\pi]$. For every $\displaystyle n$, you have $\displaystyle \lim_{x\to 0^+} f_n(x)=1$ like I said. However, you can prove that this, together with the uniform convergence, implies $\displaystyle \lim_{x\to 0^+}f(x)=1$, where $\displaystyle f$ is the limit of $\displaystyle f_n$ (write $\displaystyle |f(x)-1|\leq |f_n(x)-f(x)|+|f_n(x)-1|<2\epsilon$ for suitable choices of large $\displaystyle n$ and small $\displaystyle x$...). In the present case, $\displaystyle f(x)=0$ for $\displaystyle 0<x<\pi$, hence clearly the previous limit is wrong: this is a contradiction.