# Find the limit of the given function, 0*∞ case

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.

What is my mistake?

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,

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

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 $$\log (1+\frac 1 x)$$.

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 $$\infty$$.

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.