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Exercise 7.5 · Q11

Q.Evaluate: displaystylelimxto0+(cosx)1/x2\\displaystyle\\lim_{x\\to0^{+}}(\\cos x)^{1/x^2}

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Log-transform the 1∞1^{\infty} form to a 00\tfrac00 ratio, apply l'Hôpital twice, and exponentiate back.

Step 1. Let g(x)=(cos⁡x)1/x2g(x)=(\cos x)^{1/x^2} and take logarithms.

log⁡g(x)=log⁡(cos⁡x)x2\log g(x)=\dfrac{\log(\cos x)}{x^2}. As x→0+x\to0^+: numerator →log⁡1=0\to\log1=0, denominator →0\to0 — a 00\tfrac00 form.

Step 2. Apply l'Hôpital.

lim⁡x→0+log⁡(cos⁡x)x2=lim⁡x→0+−sin⁡x/cos⁡x2x=lim⁡x→0+(−tan⁡x2x).\lim_{x\to0^+}\frac{\log(\cos x)}{x^2}=\lim_{x\to0^+}\frac{-\sin x/\cos x}{2x}=\lim_{x\to0^+}\left(-\frac{\tan x}{2x}\right).

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

Step 3. Apply l'Hôpital again. …

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