Question

$\underset{x\to 0}{lim}\frac{4^x-2^{x+1}+1}{x\tan x}$.

Answer 1

$L=\underset{x\to 0}{lim}\frac{4^x-2^{x+1}+1}{xtanx}$

Since the limit is of the indeterminate form $\left(\frac{\to 0}{\to 0}\right)$

Ln rule can be applied

$\Rightarrow L=\underset{x\to 0}{lim}\frac{4^{x}ln4-2^{x+1}ln2}{tanx+xse{c}^{2}x}\left(\frac{\to 0}{\to 0}\right)$

Again applying LH rule to get :-

$\Rightarrow L=\underset{x\to 0}{lim}\frac{4^x\left(ln4\right)\left(ln4\right)-2^{x+1}\left(ln2\right)\left(ln2\right)}{se{c}^{2}x+se{c}^{2}x+x\left(2secx\right)\left(secxtanx\right)}$

$\Rightarrow L=\frac{4^0(ln4{)}^2-2^1(ln2{)}^2}{1+1}$

$L=\frac{(ln4{)}^2-2(ln2{)}^2}{2}=\frac{(ln4{)}^2-ln4.ln2}{2}$

$L=\frac{ln4}{2}(ln4-ln2)$

$L=ln2\left(ln2\right)$

$L=(ln2{)}^2$