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NCERT Exemplar · Q24

Q.Evaluate lim⁡x→asin⁡x−sin⁡ax−a\lim_{x \to a} \dfrac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}.

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Direct substitution gives 00\frac{0}{0}. Rewrite sin⁡x−sin⁡a\sin x-\sin a with the sum-to-product identity and rationalize x−a\sqrt{x}-\sqrt{a}, then use lim⁡θ→0sin⁡θθ=1\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1. The limit is 2acos⁡a2\sqrt{a}\cos a.

Step 1 — Check the form. Putting x=ax=a gives numerator sin⁡a−sin⁡a=0\sin a-\sin a=0 and denominator a−a=0\sqrt{a}-\sqrt{a}=0, the indeterminate form 00\frac{0}{0}. We must simplify before substituting.

Step 2 — Rewrite the numerator (sum-to-product).

sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\sin C-\sin D=2\cos\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right)

sin⁡x−sin⁡a=2cos⁡(x+a2)sin⁡(x−a2).\sin x-\sin a=2\cos\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right).

Step 3 — Rationalize the denominator. Multiply by the conjugate:

x−a=(x−a)(x+a)x+a=x−ax+a.\sqrt{x}-\sqrt{a}=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}=\frac{x-a}{\sqrt{x}+\sqrt{a}}.

Step 4 — Assemble.

lim⁡x→a2cos⁡(x+a2)sin⁡(x−a2)x−ax+a=lim⁡x→a2cos⁡(x+a2)(x+a)⋅sin⁡(x−a2)x−a.\lim_{x\to a}\frac{2\cos\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)}{\dfrac{x-a}{\sqrt{x}+\sqrt{a}}}=\lim_{x\to a}2\cos\left(\frac{x+a}{2}\right)(\sqrt{x}+\sqrt{a})\cdot\frac{\sin\left(\frac{x-a}{2}\right)}{x-a}. …

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