Wrong way of using the squeeze rule

Question

In the first part of the question, I was asked to prove $\lim_{x\to ∞} f(x)={1\over 2}$ using the squeeze rule. But I did it in a much different way than "Student A" and I fail understand what exactly he did, and where he did wrong.. any thoughts?

Answer 1

The fallacy is that the number of terms is neither constant nor bounded, so a proof by induction on $n$ that $\lim_{x\to c}\sum_{i=1}^nf_i(x)=\sum_{i=1}^n\lim_{x\to c}f_i(x)$ doesn't extend in the way Student A hoped. The correct way uses $f(x)=\tfrac{\lfloor x\rfloor(\lfloor x\rfloor+1)}{2x^2}\in\left(\tfrac{x(x-1)}{2x^2},\,\tfrac{x(x+1)}{2x^2}\right]$.

Answer 2

The problem lies in the number of terms. We have $$\lim_{x\to\infty}f(x) = \frac{1}{x^2} + \cdots + \frac{\lfloor{x}\rfloor}{x^2} \leq \lim_{x\to\infty} ( \frac{1}{x} + \frac{1}{x} + \cdots + \frac{1}{x}),$$ where the number of terms in the parenthesis is $\lfloor{x}\rfloor$.

This is equal to $$\lim_{x\to\infty}\sum_{i=1}^{i=\lfloor{x}\rfloor} \frac{1}{x},$$ and the lim cannot go through the sum, since the sum itself is not bounded and depends on $x$. You can still compute the limit and find that it is equal to $$ \lim_{x\to\infty}\sum_{i=1}^{i=\lfloor{x}\rfloor} \frac{1}{x} = \lim_{x\to\infty}\frac{\lfloor{x}\rfloor}{x} = 1. $$