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\( \lim _{x \rightarrow 0} \frac{\ln \cos x}{\sqrt[4]{1+x^{2}}-1} \) |
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Answer» \(\lim\limits_{x\to 0} \frac{ln(cos\, x)}{(1+ x^2)^{\frac14} - 1}\) (\(\frac00\) - case) \(= \lim\limits_{x\to 0} \frac{-sin\,x}{cos\, x\left(\frac14 (1 + x^2) ^{-\frac34}.2x\right)}\) (By using D.L.H. Rule) \(=\frac12 \lim\limits _{x \to 0} \frac{-sin\, x}{x} . \frac{1}{cos\,x}.(1 + x^2)^{\frac34}\) \(= \frac{-1} 2 \left(\lim\limits_{x\to 0} \frac{sin\, x} x\right) \left( \lim\limits_{x\to 0} \frac1{cos\,x}\right) \left(\lim\limits_{x\to 0} (1 + x^2)^{\frac34}\right)\) \( = \frac{-1}{2} \times 1 \times 1\times1\) \(\left(\because \lim\limits_{x \to 0} \frac{sin\, x}{x} = 1\right)\) \(= \frac{-1}{2}\) |
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