The value oflimh-0In1-2h-2In1-h-h2is

Lim_h→0In(1+2h) 2In(1+h)/h^2 =lim_h→0In(1+h)^2 In(1+2h)/h^2 =lim_h→0In ( (1+h)^2/1+2h )/h^2=lim_h→0In ( (1+h^2)/1+2h )/ ( h^2/1+2h )(1+2h) =lim_h→0In ( 1+h^2 ...

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Byju's Answer

Standard XIII

Mathematics

Standard Limits to Remove Indeterminate Form

The value ofl...

Question

$The value of {lim}_{h \to 0} \frac{I n (1 + 2 h) - 2 I n (1 + h)}{h^{2}} i s$

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Solution

The correct option is B
-1

${lim}_{h \to 0} \frac{I n (1 + 2 h) - 2 I n (1 + h)}{h^{2}} = {lim}_{h \to 0} \frac{I n (1 + h)^{2} - I n (1 + 2 h)}{h^{2}} = {lim}_{h \to 0} {\frac{I n (\frac{(1 + h)^{2}}{1 + 2 h})}{h}}^{2} = {lim}_{h \to 0} \frac{I n (\frac{(1 + h^{2})}{1 + 2 h})}{(\frac{h^{2}}{1 + 2 h}) (1 + 2 h)} = {lim}_{h \to 0} \frac{I n (1 + \frac{h^{2}}{1 + 2 h})}{(\frac{h^{2}}{.} 1 + 2 h)} \frac{1}{1 + 2 h} = - 1 [U s i n g {lim}_{x \to 0} \frac{l o g 1 + x}{x} = 1]$

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The value oflimh→0In(1+2h)−2In(1+h)h2is

The correct option is B -1 limh→0In(1+2h)−2In(1+h)h2=limh→0In(1+h)2−In(1+2h)h2=limh→0In((1+h)21+2h)h2=limh→0In((1+h2)1+2h)(h21+2h)(1+2h)=limh→0In(1+h21+2h)(h2.1+2h)11+2h=−1 [Usinglimx→0log1+xx=1]

$The value of {lim}_{h \to 0} \frac{I n (1 + 2 h) - 2 I n (1 + h)}{h^{2}} i s$