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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 ?

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    Hint:another way ,take $$x=1+a $$so $ a \to 0$ $$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\\ \lim_{a\to 0} \frac{\tan(\pi (a+1))}{(a+1)^2-\sqrt{a+1}}=\\ \lim_{a\to 0} \frac{\tan(\pi a)}{(a+1)^2-\sqrt{a+1}}=\\ \lim_{a\to 0} \frac{\pi a}{(a+1)^2-\sqrt{a+1}}=\\$$ – 2017-02-11

3 Answers 3

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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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    why $\tan (\pi x)=\tan(\pi x-\pi)$ – 2017-02-11
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    @Almot1960 $\tan x $ is periodic, with a period of $\pi$. – 2017-02-11
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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 $y\ne1$

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