Let $f \in C[0,\infty)$ and $\lim_{x\to \infty} f(x) = 1$. Prove: $$\lim_{x \to \infty} \int_{x}^{x+1} f(t) dt = 1$$
The claim is intuitive. That as $x \to \infty$ we get a $1 \times 1$ box underneath f. But here's my attempt...
It is equivalent to prove that
$$ \lim_{x \to \infty} \lim_{||P|| \to 0} \sum_{i = 1}^{n}f(t_i)(x_i - x_{i-1}) = 1$$ Where $||P||$ is the norm (or mesh) of the partition associated with the Riemann sum and $t_i$ is the "tagged" point on $[x_i, x_{i-1}]$
Proof:
$\text{Define: }x = x_0< x_1< \cdots< x_n = x+1$
$$\lim_{x \to \infty} \lim_{||P|| \to 0} \sum_{i = 1}^{n}f(t_i)(x_i - x_{i-1}) = \lim_{||P|| \to 0} \sum_{i = 1}^{n} \lim_{x \to \infty} f(t_i)(x_i - x_{i-1}) \\ = \lim_{||P|| \to 0} 1\cdot (x_1 - x_0) + 1\cdot (x_2- x_1)+\cdots+1\cdot(x_n - x_{n-1}) = \lim_{||P|| \to 0} (x+1) - x = 1 \,\, _{_\square}$$
I'm just unsure about passing that limit to the summation. Is that okay, and when is it not?
The result follows simply from the fact that $1 - \epsilon < f(x) < 1 + \epsilon$ for all sufficiently large $x$. Of course, this approach invokes a theorem stating that $f \geqslant g \implies \int f \geqslant \int g$.
You still can prove this directly (as you intended) using partitions and considering the two limits.
Since $f(x) \to 1$ as $x \to \infty$, for any $\epsilon >0$ there exists $K > 0$ such that $1 - \epsilon/2 < f(x) < 1 + \epsilon/2$ for all $x > K$. Fixing $x$ there is a partition $P_x = (x_0,x_1, \ldots, x_n)$ of $[x,x+1]$ such that for any Riemann sum $S(P_x,f)$ we have
$$S(P_x,f) - \epsilon/2 < \int_x^{x+1} f(t) \, dt < S(P_x,f) + \epsilon/2.$$
Note that for some choice of tags $\{t_j : 1 \leqslant j \leqslant n\}$
$$S(P_x,f) = \sum_{j=1}^nf(t_j)\, (x_j - x_{j-1}).$$
Since $t_j > K$ we have $1 - \epsilon/2 < f(t_j) < 1 + \epsilon/2$ for all $j$ and it follows that
$$1 - \epsilon/2 = \sum_{j=1}^n (1-\epsilon/2)(x_j- x_{j-1}) < \sum_{j=1}^n f(t_j)(x_j- x_{j-1}) \\ <\sum_{j=1}^n (1+\epsilon/2)(x_j- x_{j-1}) = 1 + \epsilon/2.$$
Thus, for all $x > K$ we have $1 - \epsilon/2 < S(P_x,f) < 1 + \epsilon/2$ and
$$1 - \epsilon < \int_x^{x+1} f(t) \, dt < 1 + \epsilon,$$
proving that
$$\lim_{x \to \infty} \int_x^{x+1} f(t) \, dt = 1.$$