Where is the given function discontinuous?
$$f(x)= \begin{cases} x^2&\quad\text{if }x < -1\\ \sqrt{x + 4}&\quad\text{if }−1 ≤ x ≤ 0\\ \sin(2x)/x&\quad\text{if }x > 0\\ \end{cases}$$
I have checked discontinuity with $x=0$ (continuous), and also with $x=-1$ (discontinuous).
Is my solution true? Is the point $x=-1$ is the only point where this function is discontinuous?
Yes that is correct. From one side the graph approaches $1$ but the other side approaches $\sqrt{3} \neq 1$. Since the limits disagree from both sides we have a jump discontinuity.
As $f(x)$ is a picewise function, in order to check its discontinuity, you should check the sets the function is defined by.
So: if $x < -1$, the function is continuous as it is defined as $x^2$, which is continuous because it is a polynomial.
If $x=-1$, we've got that $$\lim_{x \to -1^-} f(x) = 1 \neq \sqrt3 = \lim_{x \to -1^+} f(x)$$ so, directly, $f$ is discontinuous in $x=-1$.
If $-1 < x < 0$, $f$ is continuous because $\sqrt{x + 4}$ is continous in its domain (you can prove it).
If $x = 0$, we see that $$\lim_{x \to 0^-} f(x) = \sqrt{0 + 4} = 2$$ and $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sin(2x)/x = \lim_{x \to 0^+} 2\cos(2x) / 1 = 2\cos(0) = 2$$ (we have applied L'Hôpital's rule).
So, as $$f(0) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$$ we have proved that $f$ is continuous in $x=0$.
And, finally, if $x > 0$, the function is defined by $\sin(2x)/x$, which isn't defined if $x=0$. As we are seeing the behaivour of $f$ in for values of $x$ that verify $x > 0$, the function will be continuous in this part of the function.
So, where is the function discontinuous? Only in $x = -1$.
