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lim ln cot (pi/4 - k1x)tan k2x = 1.
x→0
THEN FIND RELATION BETWEEN k1 and k2.
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4 Answers
66Rewrite the given limit as
\lim_{x\to 0}\dfrac{\ln(1+\{\cot(\pi/4-k_1x)-1\})}{\cot(\pi/4-k_1x)-1} \cdot \dfrac{\cot(\pi/4-k_1x)-1}{\tan k_2x}
The first term has a limit 1. The second term is same as
\dfrac{2\cot k_1x}{(1-\cot k_1x)\tan k_2x}
=\dfrac{2\frac{\cot k_1x}{k_1x}\,k_1x}{(1-\cot k_1x)\frac{\tan k_2x}{k_2x}\,k_2x}
whose limit is \dfrac{2k_1}{k_2}
Hence we have 2k_1=k_2
1\ln{ \frac{1+\cot k_1x}{1-\cot k_1x} }= \ln{ \left( 1+\cot k_1x \right)}-\ln{\left( 1-\cot k_1x \right)}\\=2(\cot k_1x + \frac{\left( \cot k_1x\right)^3}{3}+.........\infty) \\ hence \\ \\ \frac{\ln{ \frac{1+\cot k_1x}{1-\cot k_1x} }}{\tan k_2x}=\frac{2\cot k_1 x \left( 1+\frac{\left( \cot k_1x\right)^2}{3}+.......\infty \right)}{k_2x\left(1+\frac{(k_2x)^2}{3}+....\infty \right)}\\ \lim_{x\rightarrow 0}\frac{2k_1\cot k_1 x \left(1+ \frac{\left( \cot k_1x\right)^2}{3}+.......\infty \right)}{k_2(k_1x)\left(1+\frac{(k_2x)^2}{3}+....\infty \right)}=\frac{2k_1}{k_2} \ \ \ \left(\because \lim_{x\rightarrow 0}\frac{\cot k_1x}{k_1x}=1 \right)