Check monotonicity using derivatives Depending on the $ c \in \mathbb{R}, c \not= 0$ 
Check monotonicity (increasing or non-decreasing) for $ x_{n}=(1+\frac{c}{n})^n$
I tired intruduce function $f(1/n)=(1+cn)^{\frac{1}{n}}$ and sign the derivative depends on $$\frac{\large{\frac{c n}{1+c n}}-\log(1+c n)}{n^2}$$
but dont know what to do next.
 A: Hint: Consider the function $ f(x)=(1+\frac{c}{x})^{x},\, \forall x > 0 $ and use the first derivative test. 
A: Let $x_n=\left(1+\dfrac{c}{n} \right)^n.$ Then 
$$
\ln {x_n}=n \ln{ \left(1+\dfrac{c}{n} \right)}=c\dfrac{\ln{ \left(1+\dfrac{c}{n} \right)}}{\dfrac{c}{n}}.
$$
 Consider function 
$$
g(t)=c \,\dfrac{\ln{ \left(1+t \right)}}{t}.$$
Its derivative is
$$
g'(t)=c\, \dfrac{\dfrac{t}{t+1}-\ln(1+t)}{t^2}.
$$
Denote $f(t)=\ln(1+t)$ and apply the mean value (Lagrange's) theorem to $f$  on $[0,\; t], \quad (t>0):$
$$f(t)-f(0)=\ln(1+t)-0=f'(\xi)(t-0)=\dfrac{t}{1+\xi}>\dfrac{t}{1+t},$$
since $0<\xi<t.$ Therefore, $$\ln(1+t)>\dfrac{t}{1+t},$$ 
so
$$
\begin{cases}
g'(t)<0, & c > 0, \\
 g'(t)>0, & c < 0.
\end{cases}
$$
From this we conclude that $g(t)$ decreases for $c>0$ and increases if $c < 0.$
Put $t=\dfrac{c}{n}.$
1.  $ c > 0.$ Then $t$ decreases if $n$ increases, hence $g\left(\dfrac{c}{n} \right)=\ln{x_n}$ is increasing $ \; \Rightarrow  x_n$ increases.
2. $ c < 0.$ Then $t$ increases if $n$ increases,  hence $ g \left(\dfrac{c}{n} \right)=\ln{x_n}$ is increasing $\; \Rightarrow x_n$ increases. 
