why $\lim_{x\to+∞}\dfrac{(1+x)^{3/2}}{x^{1/2}} -x\\$ \begin{align}
& \lim_{x\to+∞}\frac{(1+x)^{3/2}}{x^{1/2}} -x\\[8pt]
= {} &\lim_{x\to+∞} \dfrac{(1+(1/x))^{3/2}-1}{1/x} \\[8pt]
= {} & \lim_{x\to+∞} \dfrac{\frac{3}{2}\cdot\frac{1}{x}}{\frac{1}{x}}
\end{align}
Detailed picture with clearer specifications
 A: The manipulations may be:
$${(1+x)^\frac32 \over \sqrt x }-x = {(1+x)^\frac32\over x^\frac12 } - {x^\frac32\over x^\frac12}$$
$$= {x^\frac32\over x^\frac12}\left(\left(1+\frac 1x\right)^\frac32 - 1\right)$$
Using Newton’s general binomial theorem we have $\left(1+\frac 1x\right)^\frac32 = 1 + \binom{3/2}1 \frac 1x + \binom{3/2}2 \frac 1{x^2}\dots$, and since terms including higher powers of $\frac 1x$ are negligibly small ($\to 0$) we may write it as simply $1+\frac 3{2x}$. 
Now, we have $$x^{\frac 32 - \frac 12}\cdot\left(1+\frac 3{2x} -1\right)$$
$$=x\cdot \frac 32\cdot \frac 1x=\boxed{\frac 32\cdot \frac 1x\over \frac 1x}$$
A: Binomial series  has been used here $$(1+h)^n=1+\binom n1h+O(h^2)$$
Alternatively set $\dfrac1x=h$ to find $$F=\lim_{h\to0}\dfrac{(1+h)^{3/2}-1}h$$
Choose $\sqrt[3]{1+h}=u$
$$F=\lim_{u\to1}\dfrac{u^3-1}{u^2-1}=?$$
A: \begin{align}
&\lim_{x\to+∞}\dfrac{(1+x)^{(\frac{3}{2})}}{x^{(1/2)}} -x\\[8pt]
& \lim_{x\to+∞}\dfrac{(1+x)^{(\frac{3}{2})}(x^{-3/2})}{x^{(1/2)}x^{-3/2}} - \frac {1}{1/x} \\[8pt]
= {} & \lim_{x\to+∞} \dfrac{(1+(1/x))^{\frac{3}{2}}-1}{1/x}
\end{align}
Apply L'Hospital Rule and you get the result.
A: Let $t=\frac1x \to 0$ and $f(t)=(1+t)^{3/2}$ then by the definition of derivative
$$\lim_{x\to+∞} \dfrac{(1+(1/x))^{3/2}-1}{1/x}=\lim_{t\to 0} \dfrac{(1+t)^{3/2}-1}{t-0}=f'(0)=\frac32$$
