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Find:\( \lim _{x \rightarrow 0} \frac{x\left(5^{x}-2^{x}\right)}{\cos 5 x-\cos 3 x} \) |
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Answer» \(\lim\limits_{x\to0}\frac{x(5^x-2^x)}{cos5x - cos3x}\) (0/0 type) = \(\lim\limits_{x\to0}\frac{(5^x-2^x)+(5^xlog5-2^x log2)x}{-5sin5x+3sin3x}\) (By using D.L.H. Rule) = \(\lim\limits_{x\to0}\frac{(5^x-2^x)+(5^xlog5-2^x log2)x}{3(3x)-5(5x)}\) (If θ is very small then sin θ = θ) = - \(\lim\limits_{x\to0}\frac{(5^x-2^x)+(5^xlog5-2^x log2)x}{16x}\) = -\((\lim\limits_{x\to0}\frac{(5^x-2^x)}{16x}+\lim\limits_{x \to 0}\frac{5^xlog5-2^xlog2}{16})\) = - \((\lim\limits_{x \to 0}\frac{5^xlog5-2^xlog2}{16}+\frac{log5-log2}{16})\) (\(\because\) ao = 1 & by D.L.H. Rule) = -\((\frac{log5/2}{16}+\frac{5/2}{16})\) = -2/16 log (5/2) = -1/8 log(5/2) Hence, \(\lim\limits_{x \to 0}\frac{x(5^x-2^x)}{cos5x-cos3x}=-\frac{1}8log\frac52\) |
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