lim _x arrow 1 frac2^2 x-2-2^x+1sin^2(x-1)=

Correct answer: (B) (log 2)^2. lim _x arrow 1 frac2^2 x-2-2^x+1sin^2(x-1)=lim _x arrow 1 frac (2^x-1-1 )^2sin^2(x-1)=lim _x arrow 1 fracfrac (2^x-1-1 )^2(x-1)^2(sin^2(x-1))/((x-1)^2) =((log 2)^2)/(1^2)=(log 2)^2

Q88 / 150 · MHT CET 2022 (08 Aug Shift 1) · Mathematics

$lim_{x \to 1} \frac{2 ^{2 x - 2} - 2 ^{x} + 1}{s i n ^{2} ( x - 1 )} =$

A
$\frac{1}{2} (lo g 2)^{2}$
Correct answer
B
$(lo g 2)^{2}$
C
$2 lo g 2$
D
$2 (lo g 2)^{2}$

Solution

x \to 1 lim \frac{2 ^{2 x - 2} - 2 ^{x} + 1}{sin ^{2} ( x - 1 )} = x \to 1 lim \frac{( 2 ^{x - 1} - 1 ) ^{2}}{sin ^{2} ( x - 1 )} = x \to 1 lim \frac{\frac{( 2 ^{x - 1} - 1 ) ^{2}}{( x - 1 ) ^{2}}}{\frac{s i n ^{2} ( x - 1 )}{( x - 1 ) ^{2}}} = \frac{( lo g 2 ) ^{2}}{1 ^{2}} = (lo g 2)^{2}

Hence, the correct answer is option: 'B'.

limx→1​sin2(x−1)22x−2−2x+1​=

Solution

Relevant Questions

$lim_{x \to 1} \frac{2 ^{2 x - 2} - 2 ^{x} + 1}{s i n ^{2} ( x - 1 )} =$