If limx→0f(x)x2=a and limx→0f(1−cosx)g(x)sin2x=b (where b≠0), then limx→0g(1−cos2x)x4=

limx→0 f(1−cos⁡x)g(x)sin2⁡x=limx→0 f2sin2⋅x2g(x)2sin2⋅x2×2sin2⋅x224sin2⁡(x)2cos2⋅x2=limx→0 ag(x)×tan2⋅x2alimx→0 x22g(x)×tan2⋅x2x22 =alimx→0 x24g(x)∴ limx→0 x2g(x)=4balimx→0 g2sin2xx4limx→0 g2sin2⁡x2sin2⁡x2×2sin2⁡x2x4=a4b×4=ab

If limx→0f(x)x2=a and limx→0f(1−cosx)g(x)sin2x=b (where b≠0), then limx→0g(1−cos2x)x4=

limx→0 f(1−cos⁡x)g(x)sin2⁡x=limx→0 f2sin2⋅x2g(x)2sin2⋅x2×2sin2⋅x224sin2⁡(x)2cos2⋅x2=limx→0 ag(x)×tan2⋅x2alimx→0 x22g(x)×tan2⋅x2x22 =alimx→0 x24g(x)∴ limx→0 x2g(x)=4balimx→0 g2sin2xx4limx→0 g2sin2⁡x2sin2⁡x2×2sin2⁡x2x4=a4b×4=ab

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\lim_{x \to 0} \frac{f (x)}{x^{2}} = a$ and $\lim_{x \to 0} \frac{f (1 - \cos x)}{g (x) \sin^{2} x} = b (w h e r e b \neq 0),$ then $\lim_{x \to 0} \frac{g (1 - \cos 2 x)}{x^{4}}$ =

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{4 a}{b}$

$\frac{a}{4 b}$

$\frac{a}{b}$

$\frac{b}{a}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} lim_{x \to 0} \frac{f (1 - \cos x)}{g (x) \sin^{2} x} \\ = lim_{x \to 0} \frac{f (2 \sin^{2} \cdot \frac{x}{2})}{g (x) (2 \sin^{2} \cdot \frac{x}{2})} \times \frac{{(2 \sin^{2} \cdot \frac{x}{2})}^{2}}{4 (\frac{\sin^{2} (x)}{2}) (\cos^{2} \cdot \frac{x}{2})} \\ = lim_{x \to 0} \frac{a}{g (x)} \times \tan^{2} \cdot \frac{x}{2} \\ a lim_{x \to 0} \frac{{(\frac{x}{2})}^{2}}{g (x)} \times \frac{\tan^{2} \cdot \frac{x}{2}}{{(\frac{x}{2})}^{2}} \\ = a lim_{x \to 0} \frac{x^{2}}{4 g (x)} \\ ∴ lim_{x \to 0} \frac{x^{2}}{g (x)} = \frac{4 b}{a} \end{array}$

$\begin{array}{l} lim_{x \to 0} \frac{g (2 \sin^{2}) x}{x^{4}} \\ lim_{x \to 0} \frac{g (2 \sin^{2} x)}{{(2 \sin^{2} x)}^{2}} \times \frac{{(2 \sin^{2} x)}^{2}}{x^{4}} \\ = \frac{a}{4 b} \times 4 = \frac{a}{b} \end{array}$