$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$ find the limit:
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$$
my try :
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}\\=\frac{\tan(\pi x)}{x^2(1-\frac{1}{x\sqrt{x}})}\\=\lim_{x\to 1} \frac{\tan(\pi x)}{x^2}.\frac{1}{{1-\frac{1}{x\sqrt{x}}}}$$
now ?
 A: Note that 
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\lim_{x \to 1}\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \frac{\pi (x-1)}{x^2-\sqrt{x}}$$
Now note that $$\lim_{x \to 1}\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \frac{\pi (x-1)}{x^2-\sqrt{x}}=\lim_{x \to 1 }\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \frac{ \pi (\sqrt{x}+1)}{\sqrt{x}(x+\sqrt{x}+1)}$$
So we have that $$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\lim_{x \to 1 }\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \lim_{x \to 1}\frac{ \pi (\sqrt{x}+1)}{\sqrt{x}(x+\sqrt{x}+1)}=\frac{2 \pi }{3}$$
A: Set $\sqrt x=y\implies x=y^2$
$$\dfrac{\tan\pi x}{x^2-\sqrt x}=\dfrac{\tan\pi y^2}{y^4-y}=\pi\cdot\dfrac{y^2-1}{y(y-1)(y^2+y+1)}\cdot\dfrac{\tan\pi(y^2-1)}{\pi(y^2-1)}$$
Now $\dfrac{y^2-1}{y(y-1)(y^2+y+1)}=\dfrac{y+1}{y(y^2+y+1)}$ for $y\ne1$
A: Not fundamentally different, but shorter with equivalents:
Set $x=1+h$ ($h\to 0$). Then


*

*$\tan \pi x=\tan(\pi+\pi h)=\tan \pi h\sim_{0}\pi h$,

*$x^2-\sqrt x=(1+h)^2-\sqrt{1+h}=12+h+o(x)-(1+\frac 12h+o(h))=\frac32h+o(h)\sim_0\frac32h$,
so that
$$\frac{\tan \pi x}{x^2-\sqrt x}\sim_{x\to1}\frac{\pi h}{\frac32h}=\frac{2\pi}3.$$

