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

Question

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 ?

Answer 1

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 $$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\lim_{x \to 1 }\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \lim_{x \to 1}\frac{ \pi (\sqrt{x}+1)}{\sqrt{x}(x+\sqrt{x}+1)}=\frac{2 \pi }{3}$$

Answer 2

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 $y\ne1$

Answer 3

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 $$\frac{\tan \pi x}{x^2-\sqrt x}\sim_{x\to1}\frac{\pi h}{\frac32h}=\frac{2\pi}3.$$