$\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}}\left(\tan ^{2} \mathrm{x}\left(\left(2 \sin ^{2} \mathrm{x}+3 \sin \mathrm{x}+4\right)^{\frac{1}{2}}-\left(\sin ^{2} \mathrm{x}+6 \sin \mathrm{x}+2\right)^{\frac{1}{2}}\right)\right)$ is equal to

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Limits

lim_{x \to \frac{π}{2}} (tan^{2} x ((2 sin^{2} x + 3 sin x + 4)^{\frac{1}{2}} - (sin^{2} x + 6 sin x + 2)^{\frac{1}{2}}))

is equal to

Answer: (A)

x \to \frac{π}{2}

sin x \to 1

and

tan^{2} x \to \infty

. Rationalize the expression:

lim_{x \to \frac{π}{2}} tan^{2} x \cdot \frac{( 2 s i n ^{2} x + 3 s i n x + 4 ) - ( s i n ^{2} x + 6 s i n x + 2 )}{2 s i n ^{2} x + 3 s i n x + 4 + s i n ^{2} x + 6 s i n x + 2}

= lim_{x \to \frac{π}{2}} tan^{2} x \cdot \frac{s i n ^{2} x - 3 s i n x + 2}{9 + 9} = lim_{x \to \frac{π}{2}} \frac{t a n ^{2} x ( s i n x - 1 ) ( s i n x - 2 )}{6}

= \frac{1}{6} lim_{x \to \frac{π}{2}} \frac{s i n ^{2} x ( 1 - s i n x )}{c o s ^{2} x} = \frac{1}{6} lim_{x \to \frac{π}{2}} \frac{s i n ^{2} x ( 1 - s i n x )}{( 1 - s i n x ) ( 1 + s i n x )}

= \frac{1}{6} lim_{x \to \frac{π}{2}} \frac{s i n ^{2} x}{1 + s i n x} = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option