The limit of log(1+x)/x as x approaches 0 is equal to 1. The limx -> 0 (log(1 + x))/x formula is given by
$\lim\limits_{x \to 0} \dfrac{\log(1+x)}{x}=1$.
In this post, we will learn to prove the limit of log(1+x)/x when x tends to zero.
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Limx → 0 log(1 + x)/x
Prove that limx → 0 log(1 + x)/x = 1. |
Answer:
Note that the given limit can be written as
$\lim\limits_{x \to 0} \dfrac{1}{x} \log(1+x)$
= $\lim\limits_{x \to 0} \log(1+x)^{\frac{1}{x}}$ by the logarithm rules
= $\log \lim\limits_{x \to 0} (1+x)^{\frac{1}{x}}$ using the property limx→0 log(f(x)) = log limx→0 f(x).
= $\log e$ by the limit formula limx→0 (1+x)1/x = e.
= 1 if the base is e.
So the limit of log(1+x)/x is equal to 1 when x→0, provided that the logarithm base is e.
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FAQs
Q1: What is the limit of log(1+x)/x when x tends to 0?
Answer: The limit of log(1+x)/x when x tends to 0 is equal to 1, that is, limx→0 log(1 + x)/x =1.