# Epsilon-Delta definition of a limit

Why in the definition of a limit:

$$\lim_{{x \to a}} f(x) = L \quad \forall \epsilon > 0, \exists \delta > 0 \text{ , such that, } 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$$

We have:

$$0<|x - a| < \delta \implies |f(x) - L| < \epsilon$$

And not:

$$0 < |x - a| < \delta \iff |f(x) - L| < \epsilon$$

I mean, don't these two sentences ($$|x - a| < \delta$$ and $$|f(x) - L| < \epsilon$$) have double implication (equivalence) speaking rigorously and logically?

If we had a double implication, then we would be saying small changes in the output variable imply small changes in the input variable, which in general will not be the case. For example, consider $$f(x)= x^2$$. Then as $$x\to -1$$, $$f(x)$$ gets arbitrarily close to $$f(-1)=1=f(1)$$, but $$x$$ and $$1$$ remain very far away.

• Thank you. It makes perfect sense. Sep 16 at 5:45

If the function has an inverse and if the function and the inverse are continuous then it does. A function has an inverse if it is bijective. Let $$t=f(x)$$ and in case of bijection $$x=f^{-1}(t)$$ and follow the same epsilon-delta definition, You can convince yourself of the double implication.

• You have to be a little careful with this. It is true for continuous bijections between the reals, but it starts to become problematic when you generalize. For example, there is a continuous bijection from $\mathbb R$ to the nonnegative $x$-axis joined with the boundary of the square $[0,1]\times[0,1]$ in $\mathbb R^2$, but the inverse is not continuous.
– M W
Sep 16 at 5:52
• Thank you for the insight, I will change my answer.
– user834772
Sep 16 at 5:55

For the constant function $$f(x)=L$$, the condition $$|f(x)-L|<\epsilon$$ is always satisfied. If the reverse implication were true, then one would always have $$0<|x-a|<\delta$$, which would be absurd.