When does equality for $x_n < y_n \Rightarrow \lim_{n \rightarrow \infty} x_n \leq \lim_{n \rightarrow \infty} y_n$ hold?

Question

Let $(x_n)$ and $(y_n)$ be 2 sequences such that $\lim_{n \rightarrow \infty} x_n$ and $\lim_{n \rightarrow \infty} y_n$ exist. Then:

($\forall n$) $x_n < y_n \Rightarrow \lim_{n \rightarrow \infty} x_n \leq \lim_{n \rightarrow \infty} y_n$. On what condition on $(x_n)$ and $(y_n)$ does the equality hold?

Background to the question:

I am trying to prove Cauchy Schwarz inequality in integral form for real-valued functions using the Cauchy Schwarz inequality in summation form i.e. If $f,g \geq 0$ are square (Riemann) integrable, prove $(\int_a^b f(x)g(x)dx)^2 \leq (\int_a^b f(x)^2dx)(\int_a^b g(x)^2dx) $ using $(\sum u_iv_i)^2 \leq (\sum u_i^2)(\sum v_i^2) $.

Proof: If $P(x,t)$ be a tagged partition of $[a,b]$. \begin{align} \int_a^b f(x)g(x)dx &:= \lim_{||P|| \rightarrow 0} \sum_{i=1}^n f(x_i)g(x_i) \Delta x_i \\ &= \lim_{||P|| \rightarrow 0} \sum_{i=1}^n (f(x_i)\sqrt{\Delta x_i})(g(x_i)\sqrt{\Delta x_i}) \\ &\leq \lim_{||P|| \rightarrow 0} \sqrt{(\sum_{i=1}^n f(x_i)^2\Delta x_i) (\sum_{i=1}^n g(x_i)^2\Delta x_i)} \\(\text{Using C-S inequality and the limit inequality I asked above.}) \\ &= \sqrt{\lim_{||P|| \rightarrow 0} (\sum_{i=1}^n f(x_i)^2\Delta x_i) (\sum_{i=1}^n g(x_i)^2\Delta x_i)} \\ \Rightarrow (\int_a^b f(x)g(x)dx)^2 &\leq \lim_{||P|| \rightarrow 0} (\sum_{i=1}^n f(x_i)^2\Delta x_i) \lim_{||P|| \rightarrow 0}(\sum_{i=1}^n g(x_i)^2\Delta x_i) \\ &= (\int_a^b f(x)^2dx)(\int_a^b g(x)^2dx) \end{align} I want to check for what $f,g$ does the equality hold. For this I need to know when the equality to $x_n \leq y_n \Rightarrow \lim x_n \leq \lim y_n$ holds! Hence the question.

Answer 1

If $\lim_{n\rightarrow \infty} (y_n-x_n) = 0$, then the limits must be equal.

Answer 2

Let $$\lim_{n \rightarrow \infty} x_n =L$$

and $$\lim_{n \rightarrow \infty} y_n =M$$

We show that $ L\le M$

Suppose on the contrary, we have $l > M$.

Let $\epsilon = \frac {l-M}{2}$

For a large enough n = we have both $x_n>L-\epsilon=\frac {L+M}{2}$ and $y_n < M+\epsilon = \frac {L+M}{2}$$

That makes $ x_n > y_n$ which is not possible due to our assumption. contradiction.

The equality hold if $|x_n-y_n |\to 0$