I need to find and prove the possible existence of this limit :
The function $f(x)$ is defined as follows:$$f(x) = \left\{ \begin{array}{c} 1, x < -2 \\ 0, x > -2 \\ \end{array} \right. $$
Does $f(x)$ have a limit as $x → −2$ ? If it does, state the value of the limit, briefly justifying your answer. If there is no limit, briefly state why not.
I've tried using the $(\epsilon, δ)$-definition definition of the limit :
$\lim_{x\to-2} f(x) = a$ if for any $\epsilon > 0$ there exists $δ$ s.t. $|f(x) - a| < \epsilon$ for all $x$ s.t. $|x + 2| < δ$.
But I struggle because of two points :
- I am not sure how to operate on a step function such as this one and use it as part of the $(\epsilon, δ)$-definition of a limit.
- There are two limits, from both left and right which exist, but I am not sure if this answers the question "Does $f(x)$ have a limit as $x → −2 ?$"
Any help would be appreciated.
$\endgroup$ 31 Answer
$\begingroup$Assume it has a limit$$\lim_{x\to-2}f(x)=L$$
So, by sequential charactetisation of the limit, for any sequence $ (x_n) $ which goes to $ -2 $, the sequence of the images $ (f(x_n)) $ should go to $ L .$
But we have$$\lim_{n\to +\infty}f(-2-\frac 1n)=1$$and$$\lim_{n\to+\infty}f(-2+\frac 1n)=0$$
Done.
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