Question

Find the value of $\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1}{x+\tan x}$

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{8^x-4^x-2^x+1}{x+\tan x}$ $\left(\frac{0}{0}\right)$ form.

Now, applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{8^x\log 8-4^x\log 4-2^x\log 2}{1+{\sec }^{2}x}$

$=\frac{\log 8-\log 4-\log 2}{2}$

$=\frac{3\log 2-2\log 2-\log 2}{2}$

$=0$.