0

enter image description here

Can someone help me with a approach?

Not looking for a solution just for a hint

InsaneDream
  • 15
  • 1
    Are you okay with seeing that it is continuous on $(0,\pi)$? For continuity at $0$ we need to show that $$\lim_{x\to 0^+}\frac{1}{\sin(x)}-\frac{1}{x}=0$$ – Dave Jan 30 '20 at 19:05
  • @Dave I saw the graph, but since I understand the defintion of continuity but not really proofing it, how do you see it is continuously instinctively? – InsaneDream Jan 30 '20 at 19:14
  • Since $\sin(x)$ is continuous and nonzero on $(0,\pi)$, we get that $\frac{1}{\sin(x)}$ is continuous on $(0,\pi)$. Similarly for $\frac{1}{x}$, and so their difference is continuous. – Dave Jan 30 '20 at 19:34
  • @Dave isn't sin(pi)=0, hence (1/0) what gets us a case that is undefined, where is my mistake?Or is pi not included because of [? – InsaneDream Jan 30 '20 at 21:40
  • 1
    Yes $\pi$ is not included in $(0,\pi)=]0,\pi[$. – Dave Jan 30 '20 at 21:42

3 Answers3

2

The only thing you need to do is prove that $\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=0$. Bring to the common denominator and use L'Hopital's rule (twice).

  • If it wouldn't be 0 the function is not continious in [0,pi[ right? – InsaneDream Jan 30 '20 at 19:12
  • Correct, because you have defined $f(0)=0$. –  Jan 30 '20 at 19:20
  • with twice u mean for each seperately right? – InsaneDream Jan 30 '20 at 19:27
  • I mean this is a $\frac{0}{0}$ case: apply L'Hopital once, and you get another $\frac{0}{0}$ case, to which you apply L'Hopital again and this ends up with something finite (hopefully $0$!). Make sure you first bring to the common denominator! ($\frac{x-\sin x}{x\sin x}$) –  Jan 30 '20 at 19:32
  • I did it now and got 0 two times and came to the conclusion that according to L'Hospital the function is continious in 0, is this enough as an answer? – InsaneDream Jan 30 '20 at 21:23
  • Sort of ... You should say: according to L'Hopital's rule you have $\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=0$, which in turn means that $\lim_{x\to 0}f(x)=f(0)$ (because you've defined $f(0)=0$), which means that $f$ is continuous at $x=0$. –  Jan 30 '20 at 21:52
2

You could use Taylor series: ${ 1\over \sin x } - { 1\over x} = {x - \sin x \over x \sin x} = { x- (x-{x^3 \over 3!} + \cdots ) \over x^2-{x^4 \over 3!} + \cdots } = {x^3 \over x^2} {{1 \over 3!} +\cdots\over 1 -{x^2 \over 4!}+\dots}$.

copper.hat
  • 172,524
0

Hint

Both $\sin x$ and $x$ are continuous and non-zero over $(0,\pi)$. For checking $x=0$ simply use $$\sin x=x-{x^3\over 6}\quad,\quad |x|\ll1$$

an4s
  • 3,716
Mostafa Ayaz
  • 31,924