# To construct jump discontinuous function

$$f(x)=\begin{cases} \frac{\tan kx}{x}, & x<0\\ 3x + 2k^2, &x\ge 0 \end{cases}$$

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?

• Hint: $\lim_{x\to 0}\frac{\tan kx}{x} = k$ – DMcMor Jan 13 at 18:45
• I didn't quite understand your question, are you asking for values of $k$ such that it has a jump discontinuity? There are infinite $k$'s that would fit. – logichtech Jan 13 at 18:46
• Perhaps you are constructing a function without a jump discontinuity? – Joshua Wang Jan 13 at 18:48
• It seems the original question asked for three examples of such functions with jump discontinuities. – DMcMor Jan 13 at 18:55
• @logichtech yes that's the requirement. I'm unsure of any systematic method instead of just substituting random values. – solopolo Jan 13 at 19:20

$$\lim_{x \rightarrow 0^-} \frac{ \tan(kx)}{x} = \lim_{x \rightarrow 0^-} \frac{ \frac{k\sin(kx)}{\cos(kx)}}{kx} = \lim_{x \rightarrow 0^-} \frac{ \sin(kx)}{kx} \cdot \lim_{x \rightarrow 0^-}\frac{k}{\cos(kx)} = k$$
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 $$0.5, 0$$