# Evaluate limit?

• Jun 9th 2009, 07:27 AM
fardeen_gen
Evaluate limit?
Evaluate:

$\lim_{x\rightarrow 0} \frac{27^x - 9^x - 3^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}$
• Jun 9th 2009, 07:49 AM
Amer
Quote:

Originally Posted by fardeen_gen
Evaluate:

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

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

$
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$

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

$\frac{-2\sqrt{2}((3ln(3))^2-(2ln(3))^2-(ln(3))^2)}{1}=-2\sqrt{2}(5(ln(3))^2)$
• Jun 9th 2009, 07:56 AM
fardeen_gen
Spoiler:
$8\sqrt{2}(\ln 3)^2$
. Where did we go wrong?
• Jun 9th 2009, 08:01 AM
Amer
Quote:

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

$
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

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

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

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