# Use the formal definition of limit

by JavaProgrammer   Last Updated September 13, 2017 14:20 PM

$$\lim\limits_{x\to 1} (x^2+1)=2$$

How do I do this?

I know this:

$$|x^2-1|<\varepsilon$$

$$0<|x-1|<δ$$

But how do I continue from that?

Tags :

Let $\epsilon > 0$.

$$|(x^2+1) - 2| = |x^2-1| = |x-1|.|x+1|$$

If we set $\delta = \min(\epsilon, 1)$.

Then, $$|x-1| < \delta \Rightarrow 0 < x < 2 \Rightarrow 1 < x+1 < 3 \Rightarrow |x+1| < 3$$

but also:

$$| x-1| < \delta \Rightarrow |x-1||x+1| < 3|x-1| < 3\epsilon$$

Wyllich
September 13, 2017 14:05 PM