How do you simplify this expression? $$\lim_{h\to0}(\frac{x}{h(x+h+1)} + \frac{1}{x+h+1} - \frac{x}{h(x+1)})$$
I know the answer is $$\frac{1}{(1+x)^2}$$
But I can't get there

Take the lcm of the denominators and simplify: You will get
\begin{align}
\dfrac{x}{h(x+h+1)} + \dfrac{1}{x+h+1} - \dfrac{x}{h(x+1)} & = \dfrac{x(x+1) + h(x+1) - x(x+h+1)}{h(x+1)(x+h+1)}\\
& = \dfrac{x^2+x + hx + h - x^2 - hx - x}{h(x+1)(x+h+1)}\\
& = \dfrac{h}{h(x+1)(x+h+1)}\\
& = \dfrac1{(x+1)(x+h+1)}
\end{align}
Now let $h \to 0$.
