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

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?

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  • $\begingroup$ Hint: $\lim_{x\to 0}\frac{\tan kx}{x} = k$ $\endgroup$ – DMcMor Jan 13 at 18:45
  • $\begingroup$ 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. $\endgroup$ – logichtech Jan 13 at 18:46
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    $\begingroup$ Perhaps you are constructing a function without a jump discontinuity? $\endgroup$ – Joshua Wang Jan 13 at 18:48
  • $\begingroup$ It seems the original question asked for three examples of such functions with jump discontinuities. $\endgroup$ – DMcMor Jan 13 at 18:55
  • $\begingroup$ @logichtech yes that's the requirement. I'm unsure of any systematic method instead of just substituting random values. $\endgroup$ – solopolo Jan 13 at 19:20
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$$ \lim_{x \rightarrow 0^-} \frac{ \tan(kx)}{x} = \lim_{x \rightarrow 0^-} \frac{ \frac{k\sin(kx)}{\cos(kx)}}{kx} = \lim_{x \rightarrow 0^-} \frac{ \sin(kx)}{kx} \cdot \lim_{x \rightarrow 0^-}\frac{k}{\cos(kx)} = k$$

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 $0.5, 0$

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