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I need help finding limit of the following function when x approaching positive infinity. $f(x)=x\left(\left(1+\frac{1}{x}\right)^{x}-e\right)$

I thought that I can use the property of $\left(1+\frac{1}{x}\right)^{x}=e$, and calculate it as $ ∞ * (e-e)= ∞*0=0$ but according to the graph, it is not a limit.

enter image description here What is my mistake?

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Let $y = \left(1+\frac{1}{x}\right)^{x}$. Then,

$$\ln y = x\ln\left(1+\frac{1}{x}\right)=x\left(\frac{1}{x} -\frac1{2x^2}+O(\frac1{x^3})\right)= 1 -\frac1{2x}+O(\frac1{x^2}) $$

$$y = \exp\left( 1 -\frac1{2x}+O(\frac1{x^2}) \right) =e -\frac e{2x} + O(\frac1{x^2})$$

Thus, the limit is,

$$\lim_{x\to\infty}f(x) =\lim_{x\to\infty}x \left(e-\frac e{2x}+O(\frac1{x^2}) -e \right) =\lim_{x\to\infty} \left(-\frac e2 + O(\frac1x) \right)= -\frac e2 $$

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Your argument makes no sense. If $f(x)=x$ and $g(x)=\frac 1 x$ then can you say $\lim_{ x \to \infty} {f(x)g(x)}=\infty*0=0$. Isn't $f(x)g(x)=1$ for all $x$.

The limit is $-\frac e 2$ and you can prove this by using the Taylor expansion of $\log (1+\frac 1 x)$.

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Consider $2$ functions, $f(x)$ and $g(x)$, where

$$\lim_{x \to \infty} f(x) = \infty \tag{1}\label{eq1A}$$

$$\lim_{x \to \infty} g(x) = 0 \tag{2}\label{eq2A}$$

You can't, in general, say anything about the result, including if it even exists, of

$$\lim_{x \to \infty} f(x)g(x) \tag{3}\label{eq3A}$$

In particular, you can't necessarily conclude it's $0$. A simple example of this, i.e., $f(x) = x$ and $g(x) = \frac{1}{x}$, given in Kavi Rama Murthy's answer, shows the result is $1$ in that case. It would be $0$ if $g(x) = \frac{1}{x^2}$, but if $f(x) = x^3$ instead, then the result of \eqref{eq3A} would be $\infty$.

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If you want more than the limit of function$$f(x)=x\left(\left(1+\frac{1}{x}\right)^{x}-e\right)$$compose Taylor series $$A=\left(1+\frac{1}{x}\right)^{x}\implies \log(A)=x\log\left(1+\frac{1}{x}\right)=1-\frac{1}{2 x}+\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right)$$ $$A=e^{\log(A)}=e-\frac{e}{2 x}+\frac{11 e}{24 x^2}+O\left(\frac{1}{x^3}\right)$$ Finishing the calculations $$f(x)=-\frac{e}{2}+\frac{11 e}{24 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.

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