$$f(x)=\begin{cases} \frac{\tan kx}{x}, & x<0\\ 3x + 2k^2, &x\ge 0 \end{cases} $$
Hi! I'm trying to construct a function from this problem with jump discontinuity but from my knowledge the variable $x$ in the denominator with limit approaching $0$ for the left side of it would would result $x=0$ making it discontinuous with a vertical asymptote hence infinite discontinuity? How can I compute a value for $k$ to keep it as discontinuous with a jump?
$$ \lim_{x \rightarrow 0^-} \frac{ \tan(kx)}{x} = \lim_{x \rightarrow 0^-} \frac{ \frac{k\sin(kx)}{\cos(kx)}}{kx} = \lim_{x \rightarrow 0^-} \frac{ \sin(kx)}{kx} \cdot \lim_{x \rightarrow 0^-}\frac{k}{\cos(kx)} = k$$
Using the known $\frac{\sin(kx)}{kx}$ limit
Thus we need to check for what values of $k$ we have a continuity, or in other words the two functions have the same output at $x=0$:
$$ k = \lim_{x \rightarrow 0^+} 3x + 2k^2 = 2k^2 \\ k = 2k^2 \\ k = 0, \frac{1}{2}$$
Thus, if you want a function with a jump continuity you need to pick some values that are not $0.5, 0$