1
$\begingroup$

Let $f(x) = \left\{\begin{array}{ll} x + 2 & -3 < x < -2 \\ -x -2 & -2 \leq x < 0 \\ x + 2 & 0 \leq x < 1 \end{array}\right.$

I want to show that $f$ has a discontinuity at $x=0$ but is continuous at all other points in $(-3,1)$

Attempt: If $f$ was continuous at $0$ then $\lim_{x \to 0}f(x) = f(0)$. Now $f(0) = 2$. Not sure how to then proceed. Do I just then say $\lim_{x \to 0}f(x) = x+2$ (since x = 0?)

Hence it is discontinuous at $x = 0$

For continuity at all other points: let $\epsilon > 0$ be given. Then there exists $ \delta > 0$ such that $|x - y| < \delta \Rightarrow |(x+2) - (y+2)| = |x - y| < \epsilon$ (let $\delta = \epsilon$).

Also there exists $\delta > 0$ such that $|x - y| < \delta \Rightarrow |-x -2 - (-y -2)| = |y - x| = |x - y| < \epsilon$.

Hence $f$ is continuous at all other points in $(-3,1)$.

$\endgroup$
2
  • 2
    $\begingroup$ Writing, e.g, $\lim_{x\to 0}f(x)=x+2$. makes no sense. The left hand is not a function of $x$. $\endgroup$
    – lulu
    Sep 20 '21 at 14:38
  • $\begingroup$ As a suggestion: consider the two one-sided limits. In order for $f(x)$ to be continuous at $0$, both of those limits must exist and they must coincide. $\endgroup$
    – lulu
    Sep 20 '21 at 14:39
3
$\begingroup$

For $x<0$, $f(x)=-x-2$ and $$l_{x<0}=\lim_{x\rightarrow0}f(x)=-2$$

For $x\geq0$, $f(x)=x+2$ and $$l_{x\geq0}\lim_{x\rightarrow0}f(x)=2\rightarrow$$

$$l_{x<0}\neq l_{x\geq0}$$

hence, the function has a discontinuity at$0$.

$\endgroup$
2
$\begingroup$

Since $f(x)=-x-2$ when $x\in[-2,0)$, you have$$\lim_{x\to0^-}f(x)=-2\ne f(0),$$and therefore $f$ is discontinuous at $0$.

The only other point of $(-3,1)$ at which $f$ could be discontinuous is $-2$. But it is continuous at $-2$ since$$\lim_{x\to-2^-}f(x)=\lim_{x\to-2^+}f(x)=0=f(-2).$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.