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Question Number 146170 by EDWIN88 last updated on 11/Jul/21

$$\underset{x\to 2}{\lim }(x^2-4)\tan \left(\frac{\pi }{x}\right)=?$$

Answered by iloveisrael last updated on 11/Jul/21

Answered by mathmax by abdo last updated on 11/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=({\mathrm{x}}^2-4)\tan \left(\frac{\pi }{\mathrm{x}}\right)\Rightarrow \mathrm{f}\left(\mathrm{x}\right){=}_{\frac{1}{\mathrm{x}}=\mathrm{t}}(\frac{1}{{\mathrm{t}}^2}-4)\tan \left(\pi \mathrm{t}\right)$$$$=\frac{(1-{4\mathrm{t}}^2)\tan \left(\pi \mathrm{t}\right)}{{\mathrm{t}}^2}(\mathrm{t}\to \frac{1}{2})$$$${=}_{\mathrm{t}-\frac{1}{2}=\mathrm{y}}\frac{(1-4{(\mathrm{y}+\frac{1}{2})}^2)\tan \left(\pi (\mathrm{y}+\frac{1}{2})\right)}{{(\mathrm{y}+\frac{1}{2})}^2}$$$$=\frac{(1-4({\mathrm{y}}^2+\mathrm{y}+\frac{1}{4})\tan (\pi \mathrm{y}+\frac{\pi }{2})}{{(\mathrm{y}+\frac{1}{2})}^2}$$$$=\frac{4({\mathrm{y}}^2+\mathrm{y})}{{(\mathrm{y}+\frac{1}{2})}^2\tan \left(\pi \mathrm{y}\right)}(\mathrm{y}\to 0)$$$$\sim \frac{4\mathrm{y}(\mathrm{y}+1)}{\pi \mathrm{y}{(\mathrm{y}+\frac{1}{2})}^2}=\frac{4(\mathrm{y}+1)}{\pi {(\mathrm{y}+\frac{1}{2})}^2}\to \frac{16}{\pi }\Rightarrow$$$${\lim }_{\mathrm{x}\to 2}({\mathrm{x}}^2-4)\tan \left(\frac{\pi }{\mathrm{x}}\right)=\frac{16}{\pi }$$

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