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Exercise 7.5 · Q1

Q.Evaluate: displaystylelimxto0dfrac1−cosxx2\\displaystyle\\lim_{x\\to0}\\dfrac{1-\\cos x}{x^2}

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✓ Free question

Direct substitution gives 00\tfrac00; differentiate top and bottom, which is again 00\tfrac00, so apply l'Hôpital a second time.

Step 1. Check the form. At x=0x=0: 1−cos⁡0=01-\cos0=0, x2=0x^2=0 — a 00\tfrac00 form.

Step 2. Apply l'Hôpital once.

lim⁡x→01−cos⁡xx2=lim⁡x→0sin⁡x2x.\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{\sin x}{2x}.

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

Step 3. Apply l'Hôpital again.

lim⁡x→0sin⁡x2x=lim⁡x→0cos⁡x2=12.\lim_{x\to0}\frac{\sin x}{2x}=\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2}.

✓Final answer

lim⁡x→01−cos⁡xx2=12\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12.

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