If $f$ is a Riemannian integrable function on $[0,1]$, then $\forall \epsilon >0 \ \ \exists \delta>0 $ s.t. if the norm of the partitions $||P||< \delta$, then $\left|\sum\limits_{k=1} ^n f(x_k) (x_k - x_{k+1}) -L \right| < \epsilon $ and L is the number $\int_0 ^ 1 f(x) dx$.
Now assume that $f$ is a continuous function, then it is Riemannian integrable and for any sequence $\{ a_n \} $ that converges to 1, $f(a_n) \to f(1)$. Let $\{ b_n \}:=\{f(a_n) \} $, then by Cauchy limit theorem $\frac{1}{n }\sum\limits_{k=1} ^n b_k \to f(1) $.
If we let the norm $||P|| =\frac{1}{n }$ then why $L \neq f(1)$ by Cauchy limit theorem?