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$f(x) = \begin{cases} 1 & -1 \le x \le 1 \\ 10 & x \gt 1 \\ 10 & x \lt -1 \end{cases}$

g(x) = $\sin\frac 1x$

What happens at this limit? $\lim_{x\to 0}{{(f \circ g}})$

Do I have a discontinuity point? If I divide to the $0^+$ and $0^-$ I get sin(infinity). Does it even matter if the function is bounded?

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    any way as you defined you have $f\circ g = 1$ since $-1\le g\le 1$ – 2017-02-14
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    I understand now. The limit will be 1. What about the point itself, I'm assuming it's undefined and we have a removable discontinuity point. Is that right? – 2017-02-14
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    which removable point are you talking about? 0 I suppose we don't care as on the right and on the left the function is valued at one – 2017-02-16

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For $x \ne 0$ we have $ -1 \le g(x) \le 1$, hence $(f \circ g)(x)= 1$

Your turn !

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    I'm aware of this. I'm asking what happens at the limit when x tends to zero of the composed function? Because If I to the limit of $0_+$ then the value inside my sin function becomes infinity. How does this affect f(x)? – 2017-02-14
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    We have (f \circ g)(x)= 1 for all $x \ne 0$, hence $ \lim_{x\to 0+}(f \circ g)(x)=1= \lim_{x\to 0-}(f \circ g)(x)$ – 2017-02-14
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    Ahh yes you're right. But the point 0 itself is not defined in the composed function, right? This means we have a removable discontinuity at 0? – 2017-02-14