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Question: \[\underset{\mathbf{h \rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\mathbf{\log}_{\mathbf{e}}\mathbf{...

lim⁡h→0log⁡e(1+2h)−2log⁡e(1+h)h2\underset{\mathbf{h \rightarrow}\mathbf{0}}{\mathbf{\lim}}\frac{\mathbf{\log}_{\mathbf{e}}\mathbf{(}\mathbf{1 + 2}\mathbf{h}\mathbf{)}\mathbf{-}\mathbf{2}\mathbf{\log}_{\mathbf{e}}\mathbf{(}\mathbf{1 +}\mathbf{h}\mathbf{)}}{\mathbf{h}^{\mathbf{2}}}

A

–1

B

1

C

2

D

–2

Answer

–1

Explanation

Solution

lim⁡h→0log⁡e(1+2h)−2log⁡e(1+h)h2\lim_{h \rightarrow 0}\frac{\log_{e}(1 + 2h) - 2\log_{e}(1 + h)}{h^{2}}=

lim⁡x→a((2h)−(2h)22+(2h)33−.....∞)−2(h−h22+h33−......)h2\lim_{x \rightarrow a}\frac{\left( (2h) - \frac{(2h)^{2}}{2} + \frac{(2h)^{3}}{3} - .....\infty \right) - 2\left( h - \frac{h^{2}}{2} + \frac{h^{3}}{3} - ...... \right)}{h^{2}}

=lim⁡h→0−h2+2h3−....h2\lim_{h \rightarrow 0}\frac{- h^{2} + 2h^{3} - ....}{h^{2}} = lim⁡h→0h2{−1+2h−....}h2\lim_{h \rightarrow 0}\frac{h^{2}\{ - 1 + 2h - ....\}}{h^{2}}

= lim⁡h→0{−1+2h+....}=−1\lim_{h \rightarrow 0}\{ - 1 + 2h + ....\} = - 1.