<s>[INST] $\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$ 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 ?
 [/INST] 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 $$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\lim_{x \to 1 }\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \lim_{x \to 1}\frac{ \pi (\sqrt{x}+1)}{\sqrt{x}(x+\sqrt{x}+1)}=\frac{2 \pi }{3}$$
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