Discussion Forum

 If the function  $f(x)=\frac{(e^{kx}-1)\tan kx}{4x^{2}},x\neq0$

                                    =16                       x=0

  is continuous at x=0 , then k=...........

Answer:

Explanation:

 We have function,

$f(x)=\frac{(e^{kx}-1)\tan kx}{4x^{2}},x\neq0$

                                    =16                       x=0

is continuous at x=0

 $\therefore$    $\lim_{x \rightarrow 0} f(x)= f(0)$

 $\Rightarrow$    $\lim_{x \rightarrow 0} \frac{(e^{kx}-1)\tan kx}{4x^{2}}=16\left(\frac{0}{0}form\right)$

 $\Rightarrow$ $\lim_{x \rightarrow 0} \frac{(e^{kx}-1)}{kx}\left(\frac{\tan kx}{kx}\right)\times\frac{k^{2}}{4}=16$

 $\Rightarrow$    $\frac{k^{2}}{4}=16\Rightarrow k^{2}=64\Rightarrow k\pm 8$