Ratio of Logarithm and Algebra

Calculus Level 1

lim ⁡ x → 0 ln ⁡ ( 1 + x ) x = ? \large \lim_{x\to 0}\dfrac{\ln(1+x)}{x}=\, ?


The answer is 1.

RoYal Abhik by RoYal Abhik , 19, India

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1 solution

Santiago Hincapie
Mar 10, 2016

we know that ln ⁡ x = − ∑ k = 1 ∞ ( − 1 ) k ( − 1 + x ) k k \ln{x}=-\sum_{k=1}^{\infty}{\dfrac{(-1)^k(-1+x)^k}{k}} then lim ⁡ x → 0 ln ⁡ 1 + x x = − lim ⁡ x → 0 ∑ k = 1 ∞ ( − 1 ) k ( x ) k − 1 k = lim ⁡ x → 0 x 0 − x 1 + x 2 − ⋯ = 1 \lim_{x\to 0}\dfrac{\ln{1+x}}{x}=-\lim_{x\to 0}\sum_{k=1}^{\infty}{\dfrac{(-1)^k(x)^{k-1}}{k}}=\lim_{x\to 0}x^0-x^1+x^2-\dots=1

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The power series for l n ( x ) ln(x) is derived by using the above limit. Hence you can't use the series expansion to do this problem, as it leads to a logical fallacy.

A Former Brilliant Member - 5 years, 2 months ago

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is this limit the only form to calculate the series?

Santiago Hincapie - 5 years, 2 months ago

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Yeah, unless you define it that way, which is very artificial.

A Former Brilliant Member - 5 years, 2 months ago

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