Evaluate $\lim_{x\rightarrow \infty}x\cdot \ln(1+\frac{1}{x})$ without l'Hôpital's Rule This seems relatively easy to do with l'Hôpital, once you establish the following equality:
$x\cdot \ln(1+\frac{1}{x}) = \frac{\ln(1+\frac1x) }{\frac{1}{x}}$
Take the derivatives and the answer is 1.
What I'm trying to do is to solve it without using l'Hôpital, but I'm quite stuck. Any help would be appreciated, thanks.
 A: You could  use $$y-\frac12y^2\le \ln(1+y)<y$$ for $y>0$. This is from using the Taylor series as an  alternating series.
Or use the mean value theorem for the fraction:
$$
\frac{\ln(1+y)-\ln(1)}{(1+y)-1}
$$

Depending on if that was your definition of the logarithm, you could also use
$$
x\ln(1+\frac1x)=x(\ln(x+1)-\ln(x))=x\int_x^{x+1}\frac{dt}t=\int_0^1\frac{x\,ds}{x+s}=\int_0^1\frac{ds}{1+\frac{s}x}
$$
and as the integrand is bounded and continuous and the integration interval compact, you can switch limit and integral to get $\int_0^1ds=1$ as result.
A: Let $g(u)=\ln(1+u)$. 
$$\lim_{u\to 0}\frac{\ln(1+u)}{u}=\lim_{u\to 0}\frac{\ln(u+1)-\ln(1)}{u-0}=\lim_{u\to 0}\frac{g(u)-g(0)}{u-0}=g'(0)=1.$$
A: You might want to take a look at this novel one page proof: 
http://aleph0.clarku.edu/~djoyce/ma122/elimit.pdf
which explains without l'Hôpital's rule that
$$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e.$$
Now your question just involves taking the natural logarithm of both sides from above.  So your answer follows from this quite easily, and is given by $$\ln e=1.$$
