graphing $f(x)=x \ln \left(1+\frac{1}{x}\right)$ I was assigned to draw the graph of this function
$f(x)$=$x\ln(1+{1\over x})$.
When I calculate $\lim_{x\to \infty} f(x)$
I get $1$ but the teacher said it's not correct even though its graph on the internet shows that
$\lim_{x\to \infty}f(x)=1$.
Please tell me where did I  go wrong?
 A: You are right: $\lim_{x \to \infty}\log (1+\frac{1}{x}) \sim \lim_{x \to \infty}\frac{1}{x}$
EDIT: another way to see it is to use the definition of $e=\lim_{n \to \infty}(1+\frac{1}{n})^n$:
$$
\lim_{s \to \infty}s \log \bigg(1+\frac{1}{s} \bigg)=\lim_{s \to \infty}\log \bigg(1+\frac{1}{s}\bigg)^{s}=\log \lim_{s \to \infty}\bigg(1+\frac{1}{s} \bigg)^s=\log e^1=1
$$
A: First: You are right! The limit $1$ is correct! To prove $\lim_{x\rightarrow\infty} f(x) = 1$ you could use L'Hôpital's rule
It holds:
$$\lim_{x\rightarrow\infty} f(x) = \lim_{x\rightarrow\infty} \frac{\ln(1+1/x)}{1/x} = \lim_{x\rightarrow\infty}\frac{-\frac{1}{x^2+x}}{-\frac{1}{x^2}} =\lim_{x\rightarrow\infty} \frac{x^2}{x^2+x} =\lim_{x\rightarrow\infty} 1- \frac{1}{x+1} = 1$$
A: I am not sure of what you know but for small values of $x$, we can use a Taylor expansion like so : $$\ln(1+y) \approx y + \frac{y^2}2 + ...$$
Now let $\frac1x = y$, when $x\to \infty$, you have $y\to 0$ therefore you can write the following : $$\ln(1+\frac1x) \approx \frac1x + \frac{1}{2x^2} + ...$$
So when you multiply by $x$ you get : $$x\ln(1+\frac1x) \approx 1 + \frac{1}{2x} + ...$$
Where $...$ are other power of $\frac1x$ wich all tend to $0$ as $x\to\infty$.
It should give you a good idea to what the limit tends to although a little bit more work is needed.
A: 
Too long for a comment:

I think what happened is that your teacher forgot about the “$\ln$” and calculated $\displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x$, since $a\ln b=\ln(a^b)$. Obviously, that limit is e, whose natural logarithm is of course $1$. Then, when s/he heard you saying the limit is $1$, s/he thought you were making the common mistake of saying that $1^\infty=1$, when in fact it is an indeterminate form. In other words, I just think your teacher got confused.
