Question

If the function
$f\left(x\right)=\frac{{15}^x-3^x-5^x+1}{x\tan x},x\ne 0$ is continuous at $x=0$, then find $f\left(0\right)$.

Answer 1

${lim}_{x\to 0}\frac{{15}^x-5^x-3^x+1}{x\tan x}$

$={lim}_{x\to 0}\frac{5^x{3}^x-5^x-3^x+1}{x\tan x}$

$={lim}_{x\to 0}\frac{5^x(3^x-1)-1(3^x-1)}{x\tan x}$

$={lim}_{x\to 0}\frac{(3^x-1)(5^x-1)}{x\tan x}$

$={lim}_{x\to 0}\frac{5^x-1}{x}\times \frac{3^x-1}{x}\times \frac{x}{\sin x}\times \cos x$

$=\log 5.\log 3.1.1=\log 5.\log 3$

$\therefore f\left(x\right)=\{\begin{array}{c}\frac{{15}^x-5^x-3^x+1}{x\tan x},x\ne 0\\ \log 5.\log 3,x=0\end{array}$ is a continouous function.

And $f\left(0\right)=\log 5\log 3$