$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$ 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 ?

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 $y\ne1$
