1. Evaluate limit?

Evaluate:

$\displaystyle \lim_{x\rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$

2. Originally Posted by fardeen_gen
Evaluate:

$\displaystyle \lim_{x\rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$

$\displaystyle \lim_{x\rightarrow 0 } \frac{27^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+cosx}}=\frac{0}{0}=lop=$

$\displaystyle lim_{x\rightarrow 0 } \frac{2(ln(27)27^x-ln(9)9^x-ln(3)3^x)(\sqrt{cosx+1})}{-sinx}=\frac{0}{0}=lop$

$\displaystyle lim_{n\rightarrow 0 } \frac{2\sqrt{2}((ln27)^{2}27^x-(ln9)^{2}9^x-(ln3)^{2}3^x}{-cosx}=$

$\displaystyle \frac{-2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}=-2\sqrt{2}(5(ln(3))^2)$

Spoiler:
$\displaystyle 8\sqrt{2}(\ln 3)^2$
. Where did we go wrong?

4. Originally Posted by Amer
$\displaystyle \lim_{x\rightarrow 0 } \frac{27^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+cosx}}=\frac{0}{0}=lop=$

$\displaystyle lim_{x\rightarrow 0 } \frac{2(ln(27)27^x-ln(9)9^x-ln(3)3^x)(\sqrt{cosx+1})}{{\color{red}-}(-sinx)}=\frac{0}{0}=lop$ since cosx derivative is -sinx and you have negative sign so sinx should be instead of -sinx

$\displaystyle lim_{n\rightarrow 0 } \frac{2\sqrt{2}((ln27)^{2}27^x-(ln9)^{2}9^x-(ln3)^{2}3^x}{cosx}=$

$\displaystyle \frac{2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}=2\sqrt{2}({\color{red}5}(ln(3))^2)$

$\displaystyle \frac{2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}=$$\displaystyle (2\sqrt{2}((ln(3))^2(9-4-1))=2\sqrt{2}(4(ln(3))^2)$