If $f: [0,1]\to R$ is a continuous function, then prove that: $$\lim _{n \rightarrow \infty} \int_{1-\frac{1}{n}}^{1}f(x) d x=0$$ My try: Since $f$ is continuous at $x=1$
we have for a given $\epsilon >0$, $\exists, \delta >0$ such that $$|f(x)-f(1)|<\epsilon$$ when $1-\delta <x <1$
Now $\forall n \geq 1$ $$\left|\int_{1-\frac{1}{n}}^{1} f(x) d x-0\right| \leq \int_{1-\frac{1}{n}}^{1}|f(x)| d x\leqslant \int_{1-\frac{1}{n}}^{1}|f(x)-f(1)| d x+\int_{1-\frac{1}{n}}^{1}|f(1)| d x$$ $\implies$ $$\left|\int_{1-\frac{1}{n}}^{1} f(x) d x-0\right| \leq \int_{1-\frac{1}{n}}^{1} \varepsilon d x+\frac{f(1)}{n}=\frac{\varepsilon+f(1)}{n}$$ Thus we have: $$\lim _{n \rightarrow \infty} \int_{1-\frac{1}{n}}^{1}f(x) d x=0$$
Is this proof correct?
There is a minor issue with your proof. The equality
$$\int_{1-\frac{1}{n}}^{1} \varepsilon d x+\frac{f(1)}{n}=\frac{\varepsilon+f(1)}{n}$$ is not correct. However you could conclude by writing
$$\int_{1-\frac{1}{n}}^{1} \varepsilon d x+\frac{f(1)}{n}\le\varepsilon+\frac{\vert f(1) \vert}{n}$$ taking $N$ such that for $n \ge N$ you have $\frac{\vert f(1) \vert}{n} \le \varepsilon$ and say that as the absolute value of the integral is less than $2 \varepsilon$ for any $\varepsilon \gt 0$, it has to be equal to zero.
An easier way would be to notice that a continuous map on the closed segment $[0,1]$ is bounded. Therefore it exists $M \gt 0$ such that
$$\vert f(x) \vert \le M$$ for all $x \in [0,1]$. From there you have
$$\left\vert \int_{1-\frac{1}{n}}^{1}f(x) dx \right\vert \le \int_{1-\frac{1}{n}}^{1}\vert f(x) \vert dx \le \int_{1-\frac{1}{n}}^{1}M dx \le \frac{M}{n}$$ and the right end side of those inequalities converges to zero as $n \to \infty$.