Question

Lim x tends to 0 (cos ax)^(cosec^2 bx)

Answer 1

Answer:

Step-by-step explanation:

$let\;\;\lim_{x \to 0} (cosax)^{cosec^2bx} =e^ k\\\\k=\lim_{x\to 0}cosec^2bx(cosax-1)\\ \\k=\lim_{x\to0}\frac{b^2x^2}{sin^2bx}(-2sin^2(\frac{ax}{2}))\frac{2}{a^2x^2}\times(\frac{a^2}{2b^2})\\\\k=\lim_{x\to0}(-1)\frac{b^2x^2}{sin^2bx}\times\frac{sin^2(\frac{ax}{2})}{\frac{a^2x^2}{4}}\times(\frac{a^2}{2b^2})\\\\k=-\frac{a^2}{2b^2}\\\\\lim_{x \to 0} (cosax)^{cosec^2bx} =e^{-\frac{a^2}{2b^2}}$

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