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Limits: Equal Zero

JEE Maths question with a full step-by-step solution.

Question
If lim⁡x→0f(x)x2=a\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^{2}}=a and lim⁡x→0f(1−cos⁡x)g(x)sin⁡2x=b\displaystyle\lim_{x\to 0}\dfrac{f(1-\cos x)}{g(x)\sin^{2}x}=b where bb is not equal to zero then lim⁡x→0g(1−cos⁡2x)x4\displaystyle\lim_{x\to 0}\dfrac{g(1-\cos 2x)}{x^{4}} is
A4ab\dfrac{4a}{b}
Ba4b\dfrac{a}{4b}
Cab\dfrac{a}{b}correct
DNone
Solution
**Step 1: Useful identity.**
1−cos⁡θ=2sin⁡2θ21-\cos\theta=2\sin^{2}\dfrac{\theta}{2}
 **Step 2: Apply to 1−cos⁡x1-\cos x and sin⁡2x\sin^{2}x.**
1−cos⁡x=2sin⁡2x2,sin⁡2x=4sin⁡2x2cos⁡2x21-\cos x=2\sin^{2}\dfrac{x}{2},\quad \sin^{2}x=4\sin^{2}\dfrac{x}{2}\cos^{2}\dfrac{x}{2}
 **Step 3: Substitute into the second given limit.**
b=lim⁡x→0f ⁣(2sin⁡2x2)g(x)⋅4sin⁡2x2cos⁡2x2b=\lim_{x\to 0}\dfrac{f\!\left(2\sin^{2}\frac{x}{2}\right)}{g(x)\cdot 4\sin^{2}\frac{x}{2}\cos^{2}\frac{x}{2}}
 **Step 4: Multiply and divide by (2sin⁡2x2)2\left(2\sin^{2}\frac{x}{2}\right)^{2} to use the first limit.**
b=lim⁡x→0f ⁣(2sin⁡2x2)(2sin⁡2x2)2⋅4sin⁡4x2g(x)⋅4sin⁡2x2cos⁡2x2b=\lim_{x\to 0}\dfrac{f\!\left(2\sin^{2}\frac{x}{2}\right)}{\left(2\sin^{2}\frac{x}{2}\right)^{2}}\cdot\dfrac{4\sin^{4}\frac{x}{2}}{g(x)\cdot 4\sin^{2}\frac{x}{2}\cos^{2}\frac{x}{2}}
 **Step 5: First factor →a\to a (since 2sin⁡2x2→02\sin^{2}\frac{x}{2}\to 0 and lim⁡f(t)/t2=a\lim f(t)/t^{2}=a).**
b=a⋅lim⁡x→0sin⁡2x2g(x)cos⁡2x2=a⋅lim⁡x→0tan⁡2x2g(x)b=a\cdot\lim_{x\to 0}\dfrac{\sin^{2}\frac{x}{2}}{g(x)\cos^{2}\frac{x}{2}}=a\cdot\lim_{x\to 0}\dfrac{\tan^{2}\frac{x}{2}}{g(x)}
 **Step 6: Multiply and divide by x2x^{2}.**
b=a⋅lim⁡x→0tan⁡2x2(x/2)2⋅(x/2)2g(x)=a⋅1⋅14lim⁡x→0x2g(x)b=a\cdot\lim_{x\to 0}\dfrac{\tan^{2}\frac{x}{2}}{(x/2)^{2}}\cdot\dfrac{(x/2)^{2}}{g(x)}=a\cdot 1\cdot\dfrac{1}{4}\lim_{x\to 0}\dfrac{x^{2}}{g(x)}
 **Step 7: Solve.**
lim⁡x→0x2g(x)=4ba  ⟺  lim⁡x→0g(x)x2=a4b\lim_{x\to 0}\dfrac{x^{2}}{g(x)}=\dfrac{4b}{a}\;\Longleftrightarrow\;\lim_{x\to 0}\dfrac{g(x)}{x^{2}}=\dfrac{a}{4b}
 **Step 8: Now compute the required limit using 1−cos⁡2x=2sin⁡2x1-\cos 2x=2\sin^{2}x.**
lim⁡x→0g(1−cos⁡2x)x4=lim⁡x→0g(2sin⁡2x)(2sin⁡2x)2⋅(2sin⁡2x)2x4\lim_{x\to 0}\dfrac{g(1-\cos 2x)}{x^{4}}=\lim_{x\to 0}\dfrac{g(2\sin^{2}x)}{(2\sin^{2}x)^{2}}\cdot\dfrac{(2\sin^{2}x)^{2}}{x^{4}}
 **Step 9: First factor →a4b\to \dfrac{a}{4b}, second factor =4 ⁣(sin⁡xx)4 ⁣→4=4\!\left(\dfrac{\sin x}{x}\right)^{4}\!\to 4.
=a4b⋅4=ab=\dfrac{a}{4b}\cdot 4=\dfrac{a}{b}
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