If $\lim_{n\to\infty} t_n = t$, prove that $|t_n - t| <\frac{|t|}{2}$ when $n\to\infty$.

Question

If $\lim_{n\to\infty} t_n = t$, prove that $|t_n - t| <\frac{|t|}{2}$ when $n\to\infty$.

It is one of the assumption that my textbook used to prove: $\lim \frac{1}{t_n} = \frac{1}{t}$ given $\lim t_n = t$. There is no detailed explanation in the book, so I asked this question.

Answer 1

I'm not sure how formally you want to prove your result, but here is an example of how to do so.

Suppose the limit of the sequence $t_n$ is $t$. Then by definition of limits, we know that for any distance $\epsilon > 0$, we can find a positive integer $N_\epsilon$ such that the tail of the sequence is always within $\epsilon$ of the limit value $t$.

(Formally, every point in the tail $\{t_n : n > N_\epsilon\}$ has the property that $|t_n - t| < \epsilon$.)

In particular if $t\neq 0$, then we can pick $\epsilon = |t| / 2 > 0$, and we have our desired result:

There exists an integer $m$ such that for every $n>m$, $|t_n - t| < |t|/2$.

Answer 2

Using the formal definition of convergence, by saying that $\lim_{n\to\infty}t_n=t$ we are saying that, given any $\epsilon >0$, we can always find an $n$ such that for all $M>n$ we have $$|t_M-t|<\epsilon\qquad (1)$$ We are in this case just choosing to have $\epsilon=|t|/2$. However we can't say that $(1)$ with $\epsilon=|t|/2$ is true for all $M$, only for $M>n$ where $n$ varies based on $\{t_n\}$. This is provided $t\neq 0$. If $t=0$ then all that could be said would be $$|t_n-t|=|t|/2\qquad\text{as }n\to\infty$$ But not less than.