Finding $\lim_{x \to \infty} x(\ln(1+x) - \ln(x))$ without l'Hopital I solved the limit
$$\lim_{x \to \infty} x(\ln(1+x) - \ln(x))$$
by writing it as $\lim_{x \to \infty} \frac{\ln(\frac{1+x}{x})}{\frac{1}{x}}$ and applying l'Hopital rule but is it possible to solve it without using l'Hopital rule?
 A: \begin{align*}
x(\ln(1+x)-\ln(x))=x\ln\left(1+\dfrac{1}{x}\right)=\ln\left(1+\dfrac{1}{x}\right)^{x}\rightarrow\ln e=1,
\end{align*}
if you accept that $\left(1+\dfrac{1}{x}\right)^{x}\rightarrow e$ as $x\rightarrow\infty$.
A: By MVT there is $c \in [x,x+1]$ such as $\ln(1+x)-\ln(x)=1/c$ so $x(\ln(1+x)-\ln(x))=x/c$
$c \in [x, x+1]$ therefore $x/c \in [x/(x+1), 1]$.
So by the squeeze theorem lim is $1$.
A: Using your second form we can substitute $u=1/x$ such that
$$\lim_{u\to +0}\dfrac{\ln(1+u)}{u}= \lim_{u\to +0}\dfrac{\ln(1+u)-\ln 1}{u-0}.$$
Hence, it is possible to evaluate the limit as the derivative of $\ln(1+u)$ at $u=+0$.
A: $$\log(x+1)-\log(x)=\log\left(\frac{x+1}{x}\right)=\log\left(1+\frac{1}{x}\right)$$
We know that $\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4)$ (for $t \in (-1,1)$); substituting $t=\frac{1}{x}$ we get: $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2} \frac{1}{x^2}+\frac{1}{3} \frac{1}{x^3}+O\left(\frac{1}{x^4}\right)$, so:
$$x(\log(x+1)-\log(x))=x\left(\frac{1}{x}-\frac{1}{2} \frac{1}{x^2}+\frac{1}{3} \frac{1}{x^3}+O\left(\frac{1}{x^4}\right)\right)$$
$$x(\log(x+1)-\log(x))=1-\frac{1}{2} \frac{1}{x}+\frac{1}{3} \frac{1}{x^2}+O\left(\frac{1}{x^3}\right) \to 1$$
A: $\lim_{x \to \infty} x(\ln(1+x) - \ln(x))
$
$f(x)
=x(\ln(1+x) - \ln(x))
=x\int_x^{1+x} \dfrac{dt}{t}
$
so
$f(x) < x\dfrac1{x}
=1
$
(since $t > x$)
and
$f(x) > x\dfrac1{x+1}
=1-\dfrac1{x+1}
$
(since $t < x+1$)
so
$\lim_{x \to \infty} f(x)
= 1
$.
