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Exercise 7.5 · Q6

Q.Evaluate: displaystylelimxto0left(dfrac1sinx−dfrac1xright)\\displaystyle\\lim_{x\\to0}\\left(\\dfrac{1}{\\sin x}-\\dfrac1x\\right)

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An ∞−∞\infty-\infty form; combining over a common denominator gives a 00\tfrac00 form that needs two rounds of l'Hôpital.

Step 1. Combine into a single fraction.

1sin⁡x−1x=x−sin⁡xxsin⁡x,\frac{1}{\sin x}-\frac1x=\frac{x-\sin x}{x\sin x},

a 00\tfrac00 form as x→0x\to0 (numerator and denominator both →0\to0).

Step 2. Apply l'Hôpital once.

lim⁡x→0x−sin⁡xxsin⁡x=lim⁡x→01−cos⁡xsin⁡x+xcos⁡x.\lim_{x\to0}\frac{x-\sin x}{x\sin x}=\lim_{x\to0}\frac{1-\cos x}{\sin x+x\cos x}.

Still 00\tfrac00 at x=0x=0.

Step 3. Apply l'Hôpital again. …

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