Piecewise Continuity: f(x) = cos(x) if x <= 0, sin(x)/x if x > 0

[nbg273]

"Determine whether the fuction is continuous at x=0. Specify continuous: Right Continuous, or Left Continuous. If there is dicontinuity at x=0, specify whether it is removable or essential."

cos(x) if x<=0
f(x) =
sin(x)/(x) if x>0

So for it to be continuous, we want lim x-->a f(x) = f(a), right?

I then plugged in x for the function at the right intervals:

- from the left
lim cos(0) = 1
x-->0-

- from the right
lim sin(0)/(0) = 0
x-->0+

So lim f(x) != f(a), so therefore the fuction is NOT continuous at x=0, right?
x-->a

I'm trying to learn the basics on my own before I move on to more difficult questions, but want to make sure that this is correct. Is it?

Thank you.

 

[MarkFL]

We are given:

\(\displaystyle \displaystyle f(x)=\begin{cases}\cos(x), & x\le0 \\[3pt] \dfrac{\sin(x)}{x}, & 0<x \\ \end{cases}\)

Your methodology is correct, however your computation of one of the limits isn't:

\(\displaystyle f(0)=1\)

\(\displaystyle \displaystyle \lim_{x\to0^{-}}f(x)=1\)

\(\displaystyle \displaystyle \lim_{x\to0^{+}}f(x)=1\)

Recall:

\(\displaystyle \displaystyle \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1\)

 

[nbg273]

We are given:

\(\displaystyle \displaystyle f(x)=\begin{cases}\cos(x), & x\le0 \\[3pt] \dfrac{\sin(x)}{x}, & 0<x \\ \end{cases}\)

Your methodology is correct, however your computation of one of the limits isn't:

\(\displaystyle f(0)=1\)

\(\displaystyle \displaystyle \lim_{x\to0^{-}}f(x)=1\)

\(\displaystyle \displaystyle \lim_{x\to0^{+}}f(x)=1\)

Recall:

\(\displaystyle \displaystyle \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1\)

Oh ok, I see what I did wrong! Thank you!