Computing the best constant in classical Hardy's inequality

What you need isn't

$$\lim_{\alpha\searrow1/p}\,\left\lvert \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx - \int_1^\infty x^{-\alpha p }\mathrm dx\right\rvert=0$$

but

$$\lim_{\alpha\searrow1/p}\frac{\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx}{\int_1^\infty x^{-\alpha p }\mathrm dx}=1\;,$$

which is indeed the case, since as $\alpha\searrow1/p$, the integrals are more and more dominated by regions where $x^{-1}\ll x^{-\alpha}$. For arbitrary $b\gt1$ and $1/p\lt\alpha\lt1$, we have

$$ \begin{eqnarray} \int_1^\infty (x^{-\alpha})^p\mathrm dx &\gt& \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \\ &\gt& \int_b^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \\ &\gt& \int_b^\infty (x^{-\alpha}-b^{\alpha-1}x^{-\alpha})^p\mathrm dx \\ &=& (1-b^{\alpha-1})^p\int_b^\infty (x^{-\alpha})^p\mathrm dx \\ &=& (1-b^{\alpha-1})^pb^{1-\alpha p}\int_1^\infty (x^{-\alpha})^p\mathrm dx \\ &=& (b^{1/p-\alpha}-b^{1/p-1})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;. \end{eqnarray} $$

Then choosing $b=2^{1/\beta}$ with $\beta=\sqrt{\alpha-1/p}$ yields

$$\int_1^\infty (x^{-\alpha})^p\mathrm dx \gt \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \gt (2^{-\beta}-2^{(1/p-1)/\beta})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;. $$

Since $\beta\to0$ as $\alpha\searrow1/p$, the factor on the right goes to $1$, and thus

$$\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx\sim\int_1^\infty x^{-\alpha p }\mathrm dx\quad\text{as}\quad\alpha\searrow1/p$$

as required.

It's been a long time since the original post, but here is yet another way to show that $\frac{p}{p-1}$, $1<p<\infty$, is the best constant in Hardy's inequality once it has been established that $\|Hf\|_p\leq\frac{p}{p-1}\|f\|_p$ where $Hf(x)=\frac1x\int^x_0f(t)\,dt$. This follows a suggestion in Exercise 3.14 in Rudin's book of Real and Complex Analysis. For $A>1$, define $$ f_A(x)=x^{-1/p}\mathbf{1}_{(1,A]}(x)$$ A straight forward computation shows that $\|f_A\|^p_p=\log(A)$. On the other hand, the Hardy transform of $f_A$ is $$ F_A(x):=Hf_A(x)=\frac{p}{p-1}\frac{1}{x}\Big((\min(A,x))^{1-\tfrac{1}{p}}-1\Big)\mathbf{1}_{(1,\infty)}(x) $$ Then $$ \begin{align*} \|F_A\|^p_p &= \Big(\frac{p}{p-1}\Big)^p\Big( \int^A_1(x^{-1/p}-x^{-1})^p\,dx + \frac{(A^{\frac{1-p}{p}} - 1)^p}{p-1}\Big) \end{align*} $$ We now normalize $f_A$, by the factor $\frac{1}{\|f_A\|_p}$ to obtain $$\frac{\|F_A\|_p}{\|f_A\|_p} \geq \Big(\frac{p}{p-1}\Big)\left(\frac{\int^A_1(x^{-1/p}-x^{-1})^p\,dx}{\log(A)}\right)^{1/p}\xrightarrow{A\rightarrow\infty}\frac{p}{p-1}$$ From this, it follows immediately that $C:=\frac{p}{p-1}$ is the best constant in Hardy's inequality. In terms of operators, as user user345872 suggested above, the operator norm of the Hardy transform $f\mapsto Hf$ on $L_p((0,\infty),dx)$ is indeed $\frac{p}{p-1}$.