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Exercise 7.5 · Q10

Q.Evaluate: displaystylelimxtofracpi2−(sinx)tanx\\displaystyle\\lim_{x\\to\\frac{\\pi}{2}^{-}}(\\sin x)^{\\tan x}

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Take logarithms to reduce the 1∞1^{\infty} form to a 0×∞0\times\infty, then to a 00\tfrac00 ratio, apply l'Hôpital, and exponentiate.

Step 1. Let g(x)=(sin⁡x)tan⁡xg(x)=(\sin x)^{\tan x} and take logarithms.

log⁡g(x)=tan⁡xlog⁡(sin⁡x)\log g(x)=\tan x\log(\sin x). As x→π/2−x\to\pi/2^-: tan⁡x→+∞\tan x\to+\infty, log⁡(sin⁡x)→log⁡1=0\log(\sin x)\to\log1=0 — a 0×∞0\times\infty form.

Step 2. Rewrite as a 00\tfrac00 ratio.

tan⁡xlog⁡(sin⁡x)=log⁡(sin⁡x)cot⁡x.\tan x\log(\sin x)=\frac{\log(\sin x)}{\cot x}.

As x→π/2−x\to\pi/2^-, cot⁡x→0\cot x\to0: this is a 00\tfrac00 form.

Step 3. Apply l'Hôpital.

ddxlog⁡(sin⁡x)=cos⁡xsin⁡x=cot⁡x\dfrac{d}{dx}\log(\sin x)=\dfrac{\cos x}{\sin x}=\cot x; ddxcot⁡x=−csc⁡2x\dfrac{d}{dx}\cot x=-\csc^2x. …

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