0

Find a non-zero value for the constant k that makes $f(x)=\begin{Bmatrix} \dfrac{\tan(kx)}{x} ,& x<0 \\[6pt] 3x+2k^{2}, & x\geqslant 0 \end{Bmatrix}$ continous at $x=0$.

I've been trying to solve this question for a long time and still cant do it

user136877
  • 145
  • 1
  • 1
  • 6
  • Have you worked out the left and right hand limits at zero for this function? – Matthew Conroy Apr 16 '14 at 23:34
  • What have you tried? The first step is to find the limits $\lim_{x \to 0^-} \frac{\tan kx}{x}$ and $\lim_{x \to 0^+} 3x + 2k^2$. – fgp Apr 16 '14 at 23:35
  • for the left is $$ \lim_{x\to0} \frac{\tan(kx)}{x} = \lim_{x\to0} \left( \frac{k}{\cos(kx)} \cdot\frac{\sin(kx)}{kx} \right) $$ ,right hand is $2k^2$ – user136877 Apr 16 '14 at 23:38

1 Answers1

1

You know a condition for continuity at $0$ is

$$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^{-}} f(x) $$

Details:

$$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3x + 2k^2 = 2k^2 $$

$$ \lim_{x \to 0^-} \frac{ \tan(xk)}{x} = \lim_{x \to 0^-} \frac{ \tan(xk)}{x} \cdot \frac{k}{k} = k$$

since $ \lim_{\alpha \to 0 } \frac{ \tan \alpha}{\alpha} = 1 $. Hence

$$ 2k^2 = k \iff k(2k-1) = 0 $$

....