How Does Subtracting a Limit Value Affect Convergence?

[Amad27]

Prove that $\displaystyle \lim_{x\to a} f(x) = L \space \text{if and only if} \space \lim_{x\to a} [f(x) - L] = 0$ Provide a rigorous proof.

I am not sure what he has given to us.

Is $\displaystyle \lim_{x\to a} f(x) = L$ true?

So,

$|f(x) - L| < \epsilon$ for $|x - a| < \delta_1$ some $\delta_1$

$$\lim_{x\to a} f(x) - L = 0 \implies |f(x) - L| < \epsilon \space \text{such that} \space |x - a | < \delta_2$$

I feel we need to prove that $\delta_1 = \delta_2$ Can someone confirm this?

But how do we prove this?

 

[Siron]

I think your idea is correct but I would write it down as follows.

To prove (the other direction is similar):
$$\lim_{x \to a} f(x)=L \Rightarrow \lim_{ x \to a}[f(x)-L]=0$$

Proof
Let $\epsilon>0$, as $\lim_{x \to a} f(x)=L$ there exists a $\delta>0$ such that
$$0<|x-a|<\delta \Rightarrow |f(x)-L|=|[f(x)-L]-0|<\epsilon$$

But that means we've found for every $\epsilon>0$ a suitable $\delta$ such that the above statement holds, that is, $\lim_{x \to a} [f(x)-L]=0$.