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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}\!\left(\dfrac{\sin x}{x}\right)^{\!\!1/x^{2}}

e^{1/2}

1

e^{-1/6}

correct

e^{1/3}

Solution

**Step 1: Identify the form.**

\dfrac{\sin x}{x}\to 1

and

\dfrac{1}{x^{2}}\to\infty

. Form is

1^{\infty}

. **Step 2: Standard formula for

1^{\infty}

.**

\lim f(x)^{g(x)}=\exp\!\left[\lim g(x)\!\left(f(x)-1\right)\right]

**Step 3: Apply with

f(x)=\dfrac{\sin x}{x},\;g(x)=\dfrac{1}{x^{2}}

.**

L=\exp\!\left[\lim_{x\to 0}\dfrac{1}{x^{2}}\!\left(\dfrac{\sin x}{x}-1\right)\right]=\exp\!\left[\lim_{x\to 0}\dfrac{\sin x-x}{x^{3}}\right]

**Step 4: Use Taylor expansion

\sin x=x-\dfrac{x^{3}}{6}+\ldots

.**

\dfrac{\sin x-x}{x^{3}}=\dfrac{-x^{3}/6+\ldots}{x^{3}}\to -\dfrac{1}{6}

**Step 5: Substitute.**

L=e^{-1/6}

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