Find the limit of the given expression

Question

Find the limit o the expression: $$ \lim_{x\rightarrow 1}\frac{x^p-1}{x^q-1}\qquad\qquad ,>0 $$

I can do it using L hopitals rule bt is there a way to do it without differentiation??

Answer 1

We know

$$ x \to 1,\ln x=\ln(1+x-1) \to x-1 $$

Thus,when $x \to 1$

$$ x^p-1 \to \ln(x^p)=p \ln x $$

$$ x^q-1 \to \ln(x^q)=q \ln x $$

$$ L=\lim_{x \to 1} \frac{p \ln x}{q \ln x}=\frac{p}{q} $$

Answer 2

If $ p,q $ are integers, it's more simple to use that : $ x^{p}-1=\left(x-1\right)\sum\limits_{k=0}^{p-1}{x^{k}} $ and that : $ x^{q}-1=\left(x-1\right)\sum\limits_{k=0}^{q-1}{x^{k}} $.

If $ p,q $ are reals, then we'll be using the well-known limit : $ \lim\limits_{x\to 0}{\frac{\mathrm{e}^{x}-1}{x}}=1 : $ \begin{aligned}\lim_{x\to 1}{\frac{x^{p}-1}{x^{q}-1}}&=\lim_{x\to 1}{\left(\frac{p}{q}\times\frac{\mathrm{e}^{p\ln{x}}-1}{p\ln{x}}\times\frac{q\ln{x}}{\mathrm{e}^{q\ln{x}}-1}\right)}\\&=\frac{p}{q}\times 1\times 1\\ &=\frac{p}{q}\end{aligned}