I have this function:
$f(x) = \begin{cases} \tan(\frac{\pi x}{4}), & \lvert x \rvert \lt 1 \\ x, & \lvert x \rvert \ge 1 \end{cases}$
I presumed it as continuous just because I analysed the limit as the function approaches 1.
$\lim \limits_{x \to 1^+}f(x)=(1)=1$
$\lim \limits_{x \to 1^-}f(x)=\tan(\frac{\pi (1)}{4})=\tan(\frac{\pi}{4})=1$
$\lim \limits_{x \to -1^+}f(x)=(-1)=-1$
$\lim \limits_{x \to -1^-}f(x)=\tan(\frac{\pi (-1)}{4})=-\tan(\frac{\pi}{4})=-(1)=-1$
Are there any cases in which the function is discontinuous? I must not use taking derivatives or anything higher, just using basic limit definitions should be used: the problem assumes that I don't know anything from higher level calculus.
Correct, $x=\pm1$ is the part that takes work. $\tan(x)$ is continuous on (−π/2,π/2), where you are using that it is continuous on [−π/4,π/4], and $f(x)=x$ is continuous everywhere.