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Limits — JEE Maths practice question

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

Question

\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^{2}}=a

and

\displaystyle\lim_{x\to 0}\dfrac{f(1-\cos x)}{g(x)\sin^{2}x}=b

where

b

is not equal to zero then

\displaystyle\lim_{x\to 0}\dfrac{g(1-\cos 2x)}{x^{4}}

\dfrac{4a}{b}

\dfrac{a}{4b}

\dfrac{a}{b}

correct

DNone

Solution

**Step 1: Useful identity.**

1-\cos\theta=2\sin^{2}\dfrac{\theta}{2}

**Step 2: Apply to

1-\cos x

and

\sin^{2}x

.**

1-\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\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

\left(2\sin^{2}\frac{x}{2}\right)^{2}

to use the first limit.**

b=\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

\to a

(since

2\sin^{2}\frac{x}{2}\to 0

and

\lim f(t)/t^{2}=a

).**

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

x^{2}

.**

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\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=2\sin^{2}x

.**

\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

\to \dfrac{a}{4b}

, second factor

=4\!\left(\dfrac{\sin x}{x}\right)^{4}\!\to 4

=\dfrac{a}{4b}\cdot 4=\dfrac{a}{b}

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