Question

solve part 5 of this question ...​

Answer 1

Answer:

$\large\bold\red{ln(4)ln(2)}$

Step-by-step explanation:

Given,

$(v)\lim_{x\to0} \frac{ {8}^{x} - {4}^{x} - {2}^{x} + 1 }{x \tan(x) }$

We need to factorise the numerator here,

$= \lim_{x\to0} \frac{ {4}^{x} ( {2}^{x} - 1) - 1( {2}^{x} - 1)}{x \tan(x) } \\ \\ = \lim_{x\to0} \frac{( {4}^{x} - 1)( {2}^{x} - 1)}{x \tan(x) } \\ \\ = \lim_{x\to0} \frac{( {4}^{x} - 1)( {2}^{x} - 1)}{ {x}^{2} \times \frac{ \tan(x) }{x} } \\ \\ = \lim_{x\to0} \frac{( {4}^{x} - 1)}{x} \times \lim_{x\to0} \frac{( {2}^{x} - 1)}{x} \times \frac{1}{\lim_{x\to0} \frac{ \tan(x) }{x}}$

But,

we know that,

$\lim_{x\to0} \frac{ {a}^{x} - 1}{x} = ln(a) \\ \\ and \\ \\ \lim_{x\to0} \frac{ \tan(x) }{x} = 1$

Therefore,

Applying these formulas,

we get,

$= \bold{ ln(4) ln(2) }$