How to prove that if $x$ is nearly equal to 1, then $$px^p−qx^q≈(p−q)x^{p+q},$$ where $p$ and $q$ are any numbers?
My try is this:
Since $x$ is nearly equal to $1$, put $x=1+h$ or $x=1−h$, where $h$ is of very small magnitude that its square and higher powers can be neglected… Now the left hand side of the given approximation can be written as\begin{align*} &\mathrel{\phantom{=}}{} (p(1+h))p−(q(1+h))q\\ &=p(1+ph+⋯)−q(1+qh+⋯) \quad \text{(Using binomial theorem)}\\ &≈(p(1+ph))−(q(1+qh))\\ &\mathrel{\phantom{=}}{} \text{(The square and higher powers of }h\text{ were neglected)}\\ &=(p+p^2h)−q−q^2h \quad \text{(Distribution law)}\\ &=p−q+h(p^2−q^2) \quad \text{(Argument and take }h\text{ as common factor)}\\ &=p−q+h(p−q)(p+q)\\ &=(p−q)(1+h(p+q)) \quad \text{(}(p-q)\text{ common factor)}\\ &=(p−q)((1+h(p+q))\\ &\mathrel{\phantom{=}}{} \text{(Binomial theorem and }h\text{ for powers higher than }1\text{ is neglected)}\\ &=(p−q)x^{p+q}, \end{align*} which is the R.H.S.
But why only the binomial theorem of fractional and negative powers work here?
$px^p−q^x≈(p−q)x^p+q$ is not true for $x$ nearly equal to$1$, since
$px^p−q^x \to p-q$ as $x \to 1$, but $(p−q)x^p+q \to p-q+q=p$ as $x \to 1$.