Prove by the limit definition that the limit of a piecewise function is 1.

Question

Prove by definition that: $$\begin{equation*} \lim_{x \rightarrow 1} f(x)=1 \end{equation*},$$

Where,$$\begin{equation*} f(x) = \left\{ \begin{array}{rl} 4x - 3 & \text{if } x > 1,\\ 2-x^2 & \text{if } x < 1. \end{array} \right. \end{equation*}$$

I do not know how to find $\delta$, when we have piece-wise function, could anyone explain this for me please?

Answer 1

When you have a piecewise function, compute the $\delta$ "for each side" and take the minimum.

In our case, note that $|(4x - 3)- 1| < \epsilon$ if $|x - 1| < \frac{\epsilon}{4}$, so here $\delta_1 = \frac{\epsilon}{4}$ would work.

Similarly, note that $|x +1| \leq |x-1| + 2$, and therefore $|(2-x^2)-1| = |x-1||x+1| \leq (|x-1|)(|x-1|+2)$. Now, pick $\delta$ such that $\delta_2(\delta_2 + 2) < \epsilon$, and that would work.

Finally, taking the minimum of these $\delta_i$ will give the result, for then both inequalities desired for the limit to exist, would be true.

Answer 2

First of all, we need to check that there actually is a limit. We can check this using $3$ conditions:

$1)$ The left-hand limit equals the right-hand limit.
$2)$ The two-sided limit is equal to the point.

Looking at these 2 conditions, does the limit exist (both of these conditions must be satisfied). If the answer is yes, then you can just substitute $x=1$ into the limits.

Answer 3

Hint: Find $\delta_1$ for right hand limit and $\delta_2$ for left hand limit and then take $$\delta=\min\{\delta_1,\delta_2\}$$