Find $\lim_{x\to 1}\frac{p}{1-x^p}-\frac{q}{1-x^q}$ [duplicate]

Solution 1:

HINT:

Set $x=1+h$

$$\implies\dfrac p{1-x^p}=\dfrac p{1-(1+h)^p}=-\dfrac p{hp+\dfrac{p(p-1)h^2}2+\text{ terms containing } h^3}$$

$$=-\dfrac1{h+\dfrac{(p-1)h^2}2+\text{ terms containing } h^3}$$

So, the result will be $$\dfrac{p-1}2-\dfrac{q-1}2$$

Solution 2:

Rearrange:

$$\lim_\limits{x\to 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right)=\lim_\limits{x\to 1}\frac{p-q+px^q-qx^p}{1-x^p-x^q+x^{p+q}}\stackrel{l'H}=\lim_\limits{x\to 1}\frac{pqx^{q-1}-pqx^{p-1}}{-px^{p-1}-qx^{q-1}+(p+q)x^{p+q-1}}\stackrel{l'H}=\lim_\limits{x\to 1}\frac{pq(q-1)x^{q-2}-pq(p-1)x^{p-2}}{-p(p-1)x^{p-2}-q(q-1)x^{q-2}+(p+q)(p+q-1)x^{p+q-2}}=\frac{pq(p-q)}{2pq}=\frac{p-q}{2}.$$

Solution 3:

Hint: write your term in the form $$\frac{p(1-x^q)-q(1-x^p)}{(1-x^p)(1-x^q)}$$ and use L'Hospital. the result is $$\frac{1}{2}(p-q)$$