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$\lim\limits_{h\to 0}\large\frac{ln(1+2h)-2ln(1+h)}{h^2}=$

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;-1$

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$\lim\limits_{h\to 0}\large\frac{ln(1+2h)-2ln(1+h)}{h^2}=$$\lim\limits_{h\to 0}ln\bigg[\large\frac{1+2h}{-\Large\frac{1+2h+h^2}{h^2}}\bigg]$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{ln\big[1+\Large\frac{-h^2}{1+2h+h^2}\big]}{\Large\frac{-h^2}{1+2h+h^2}}\times \large\frac{-h^2}{\Large\frac{1+2h+h^2}{h^2}}$
Using $\lim\limits_{x\to 0}\large\frac{\log(1+x)}{x}$$=1$
$\Rightarrow 1.\lim\limits_{h\to 0}\large\frac{-1}{1+2h+h^2}$
$\Rightarrow -1$
Hence (d) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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