Find the limit $\lim\limits_{x\to\infty} f(x)= \lim\limits_{x \to \infty} \left(\frac{x}{x+1}\right)^x$

Let's start with finding the limit of the reciprocal of the original expression: \begin{align} & \left(1 + \frac{1}{x}\right)^x \\ \end{align} whose limit is the well-known $e$. Now you can conclude easily.

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For those who needed more details, let me put it more explictly: $$\left(\frac{x}{x + 1}\right)^x = \left(\frac{1}{1 + \frac{1}{x}}\right)^x = \frac{1}{\left(1 + \frac{1}{x}\right)^x}.$$ Now use $$\lim_{x \to \infty}\frac{h(x)}{g(x)} = \frac{\lim\limits_{x \to \infty}h(x)}{\lim\limits_{x \to \infty} g(x)}$$ when $\lim_{x \to \infty} g(x) \neq 0$. We conclude that $$\lim_{x \to \infty}\left(\frac{x}{x + 1}\right)^x = \frac{1}{\lim\limits_{x \to \infty} \left(1 + \frac{1}{x}\right)^x} = \frac{1}{e}.$$

There are a variety of approaches to evaluate the limit of interest. I thought that it would be instructive to show a way forward that uses only the squeeze theorem and standard inequalities. To that end we proceed.

I showed in THIS ANSWER and THIS ONE that the logarithm function satisfies the inequalities for $z>0$

$$\frac{z-1}{z}\le \log z \le z-1 \tag 1$$

Another way to obtain $(1)$, is to rely on the integral definition of the logarithm expressed as

$$\log z=\int_1^z \frac{1}{u}\,du$$

Now let $z=\frac{x}{x+1}$. Then, we have

$$-1=x\,\left(\frac{\left(\frac{x}{x+1}-1\right)}{\frac{x}{x+1}}\right) \le x\,\log \left(\frac{x}{x+1}\right)\le x\,\left(\frac{x}{x+1}-1\right)=-\frac{x}{x+1}$$

Using the squeeze theorem, we see that

$$\lim_{x\to \infty}x\,\log\left(\frac{x}{x+1}\right)=-1$$

Finally, using continuity of the exponential function reveals

$$\begin{align} \lim_{x\to \infty}\left(\frac{x}{x+1}\right)^x&=\lim_{x\to \infty}e^{x\log \left(\frac{x}{x+1}\right)}\\\\ &=e^{\lim_{x\to \infty}x\log \left(\frac{x}{x+1}\right)}\\\\ &=e^{-1} \end{align}$$

Another way to get it.

Consider $$A=\left(\frac{x}{x+1}\right)^x$$ and take logarithms $$\log(A)=x \log\left(\frac{x}{x+1}\right)=x \log\left(\frac{x+1-1}{x+1}\right)=x \log\left(1-\frac{1}{x+1}\right)$$ Now, remember that, for small $y$, $\log(1-y)\approx -y$. Replace $y$ by $\frac{1}{x+1}$ which makes $$\log(A)\approx -\frac{x}{x+1}=-\frac{1}{1+\frac 1x}\to -1$$ and then $A\to e^{-1}$