# Thread: Prove that h is not continuous at 0

1. ## Prove that h is not continuous at 0

Let L subset to R and define

h(x) = {sin(1/x), x not = 0
h(x) = { L, x = 0

Prove that h is not continuous at 0. (Hint: Prove by contradiction,
taking epsilon = 1/2 and use question 1)

question 1:

Suppose delta > 0.

(a) Prove that there exists a positive integer n such that

0 < 1/((4n+1)pi/2) < delta

and sin ((4n+1)pi/2) = 1

(b) Prove that there exists a positive integer m such that

0 < 1/((4m+3)pi/2) < delta

and sin ((4m+3)pi/2) = -1

Can anyone help me with this, I appreaciate it

2. Originally Posted by 450081592
Let L subset to R and define

h(x) = {sin(1/x), x not = 0
h(x) = { L, x = 0

Prove that h is not continuous at 0. (Hint: Prove by contradiction,
taking epsilon = 1/2 and use question 1)

question 1:

Suppose delta > 0.

(a) Prove that there exists a positive integer n such that

0 < 1/((4n+1)pi/2) < delta

and sin ((4n+1)pi/2) = 1

(b) Prove that there exists a positive integer m such that

0 < 1/((4m+3)pi/2) < delta

and sin ((4m+3)pi/2) = -1

Can anyone help me with this, I appreaciate it
Remember that to prove $h$ is not continous at $0$ we must show that there exist an $\epsilon >0$ such that for all $\delta >0$ there exists an $x \in \mathbb{R}$ such that $\vert x \vert < \delta$ and $\vert L - \sin (1/x) \vert \geq \epsilon$. As you're advised take $\epsilon=1/2$ and assume $L \in [0, \infty)$ then by (b) of question 1 we have an $x$ such that $0 and $\sin (1/x)=-1$, now since $L\geq 0$ we have $\vert L-\sin (1/x) \vert >1>1/2$. If $L \in (-\infty,0)$ use (a) and a similar reasoning