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$$ \lim_{x \rightarrow 0}{\frac{1}{x}-\frac{\ln (x+1)}{x^2}}$$

Can anyone help me by pointing out where I am wrong while solving this limit? The following is my attempt.

$$ \begin{aligned} & =\lim_{x \rightarrow 0}{\ln e^{1/x}-\ln(x+1)^{1/x^2}} \quad \left\{\frac{1}{x}=\ln e^{1/x}\right\} \\ & =\lim_{x \rightarrow 0} \ln{\frac{e^{1/x}}{(x+1)^{\frac{1}{x} \cdot \frac{1}{x}} }} \\ & =\lim_{x \rightarrow 0} \ln\left(\frac{e}{(x+1)^{1/x}}\right)^{1/x} \\ & =\lim_{x \rightarrow 0} \ln(1)^{1/x} \qquad \big({\text{As $(x+1)^{1/x}$ in denominator will} \\ \text{ approach $e$ as $x \rightarrow 0$ and cancel}\big)}\\\ & =\lim_{x \rightarrow 0} \ln{1} \\ & =0 \end{aligned}$$

which is the wrong answer. The correct answer is $\frac{1}{2}$.

Integreek
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    Here is a guide on mathjax for your future questions – Dstarred Dec 17 '24 at 03:07
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    This is the same mistake as in$$\lim_{x\to0}\frac xx=\lim_{x\to0}\frac 0x=0.$$ – Anne Bauval Dec 17 '24 at 04:59
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    As @Gonçalo noticed, the question in your title does not match with the one in your body. Your title is therefore misleading (and is anyway essentially a duplicate of https://math.stackexchange.com/questions/37451/how-to-calculate-lim-x-to-0-left-frac1x-frac-ln1-xx2-right?noredirect=1). Please update your title. – Anne Bauval Dec 17 '24 at 06:06
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    Kindly consider accepting an answer if you find it helpful and satisfactory. – Integreek Dec 24 '24 at 17:09

2 Answers2

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The answer is 1/2 Using the Taylor series expansion of log. taylor series of $\ln(1+x)$?

$$ f(x) = \frac{1}{x} - \frac{\log(x+1)}{x^2} $$

$$ \log(x+1) = x - \frac{x^2}{2} + O(x^3) $$

$$ f(x) = \frac{1}{x} - \frac{x - \frac{x^2}{2} + O(x^3)}{x^2} $$

$$ f(x) = \frac{1}{x} - \frac{x}{x^2} + \frac{x^2}{2x^2} - \frac{O(x^3)}{x^2} $$

$$ f(x) = \frac{1}{x} - \frac{1}{x} + \frac{1}{2} - O(x) $$

$$ f(x) = \frac{1}{2} - O(x) $$

$$ \lim_{x \to 0} f(x) = \frac{1}{2} $$

Riemann's Last Theorem 0bq.com
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    This does not answer the question, which was: "pointing out where I am wrong?". – Anne Bauval Dec 17 '24 at 04:53
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    @AnneBauval, Riemann's Last Theorem answered the question in the title. – Gonçalo Dec 17 '24 at 05:13
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    @Gonçalo The question is always in the body, not in the title: "From How to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." – Anne Bauval Dec 17 '24 at 05:58
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You have arrived at the limit $\displaystyle \lim_{x\to0}\ln\left(\frac{e}{(1+x)^\frac1x}\right)^\frac1x$ via the correct procedure. Note that it is of the indeterminate form $\vec1^{\vec\infty}$. You made a mistake in taking the limit of the base first ignoring the power which is not correct as both the base and the power are varying w.r.t $x$ simultaneously, i.e., as $x$ approaches $0$, the base gets closer and closer to $1$ but never equal to $1$ and the power gets arbitrarily large.

You can also see that if the same process is followed, any limit of this form would always be equal to $1$, which is a clear indication of this reasoning being faulty. For example:

$$\lim_{x\to0}\left(1+x\right)^\frac1x\ne\lim_{x\to0}1^\frac1x=1$$

The general procedure for solving limits of the form $\vec1^{\vec\infty}$ is

$$\lim_{x\to a}(f(x))^{g(x)}=e^{\displaystyle\lim_{x\to a}g(x)\ln f(x)}=e^{\displaystyle\lim_{x\to a}g(x)\left(f(x)-1-\frac{(f(x)-1)^2}2+\frac{(f(x)-1)^3}3\cdots\right)}$$

Usually, only the term involving $f(x)-1$ is required to solve the limit as the other terms evaluate to zero, but one should always check this.

Hence, the correct way to evaluate your limit would be

$$\require{cancel}\begin{align}\lim_{x\to0}\ln\left(\frac{e}{(1+x)^\frac1x}\right)^\frac1x&=\lim_{x\to0}\ln\left(\frac{(1+x)^\frac1x}e\right)^\frac{-1}x\\&=\lim_{x\to0}\ln\left(\frac{\cancel{e}\left(1-\frac{x}2+\frac{11x^2}{24}\cdots\right)}{\cancel{e}}\right)^\frac{-1}x\\&=-\lim_{x\to0}\frac{\left(-\frac{x}2+\frac{11x^2}{24}\cdots\right)-\overbrace{\frac{\left(-\frac{x}2+\frac{11x^2}{24}\cdots\right)^2}2\cdots}^{\text{lowest common power of }x\text{ is }2}+\overbrace{\frac{\left(-\frac{x}2+\frac{11x^2}{24}\cdots\right)^3}3\cdots}^{\text{lowest common power of }x\text{ is }3}}x\\&=\boxed{\frac12}\end{align}$$

The Taylor expansion of $(1+x)^\frac1x$ can be obtained by rewriting it as $e^{\frac{\ln(1+x)}x}$ and using the Taylor expansions of $\ln(1+x)$ and $e^x$.

Integreek
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