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Exercise 9.6 · Q19

Q.The value of lim⁡x→0sin⁡xx2\displaystyle\lim_{x\to0}\dfrac{\sin x}{\sqrt{x^2}} is

(1) 11
(2) −1-1
(3) 00
(4) ∞\infty
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Step 1. Recall x2=∣x∣\sqrt{x^2}=|x| for real xx, so the expression is sin⁡x∣x∣\dfrac{\sin x}{|x|}.

Step 2. Right-hand limit (x→0+x\to0^+): ∣x∣=x|x|=x, so sin⁡xx→1\dfrac{\sin x}{x}\to1.

Step 3. Left-hand limit (x→0−x\to0^-): ∣x∣=−x|x|=-x, so sin⁡x−x=−sin⁡xx→−1\dfrac{\sin x}{-x}=-\dfrac{\sin x}{x}\to-1. …

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