displaystyle lim x arrow 0 ( ln (2- cos 2 x)/ ln 2( sin 3 x+1)) is equal to

Question

displaystyle lim x arrow 0 ( ln (2- cos 2 x)/ ln 2( sin 3 x+1)) is equal to (A) (2/9) (B) -(2/9) (C) 0 (D) None of these

Tardigrade Admin · Accepted Answer

displaystyle lim x arrow 0 ( ln (2- cos 2 x)/ ln 2( sin 3 x+1)) = displaystyle lim x arrow 0 ( ln 1+(1- cos 2 x) / ln 2(1+ sin 3 x)) = displaystyle lim x arrow 0 (1- cos 2 x/( sin 3 x)2) = displaystyle lim x arrow 0 (2 x2/(3 x)2)=(2/9)

Q. $x \to 0 lim \frac{ln ( 2 - cos 2 x )}{ln ^{2} ( sin 3 x + 1 )}$ is equal to

$\frac{2}{9}$

$- \frac{2}{9}$

$0$

None of these

$x \to 0 lim \frac{ln ( 2 - cos 2 x )}{ln ^{2} ( sin 3 x + 1 )}$
$= x \to 0 lim \frac{ln { 1 + ( 1 - cos 2 x )}}{ln ^{2} ( 1 + sin 3 x )}$
$= x \to 0 lim \frac{1 - cos 2 x}{( sin 3 x ) ^{2}}$
$= x \to 0 lim \frac{2 x ^{2}}{( 3 x ) ^{2}} = \frac{2}{9}$

Q. x→0lim​ln2(sin3x+1)ln(2−cos2x)​ is equal to

92​

−92​

0

None of these

x→0lim​ln2(sin3x+1)ln(2−cos2x)​ =x→0lim​ln2(1+sin3x)ln{1+(1−cos2x)}​ =x→0lim​(sin3x)21−cos2x​ =x→0lim​(3x)22x2​=92​

Q. $x \to 0 lim \frac{ln ( 2 - cos 2 x )}{ln ^{2} ( sin 3 x + 1 )}$ is equal to

$\frac{2}{9}$

$- \frac{2}{9}$

$0$

$x \to 0 lim \frac{ln ( 2 - cos 2 x )}{ln ^{2} ( sin 3 x + 1 )}$
$= x \to 0 lim \frac{ln { 1 + ( 1 - cos 2 x )}}{ln ^{2} ( 1 + sin 3 x )}$
$= x \to 0 lim \frac{1 - cos 2 x}{( sin 3 x ) ^{2}}$
$= x \to 0 lim \frac{2 x ^{2}}{( 3 x ) ^{2}} = \frac{2}{9}$