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Question Number 94849 by mathocean1 last updated on 21/May/20

$$\mathrm{Calculate}\mathrm{limits}\mathrm{of}f\mathrm{at}+\mathrm{\infty };-1\mathrm{and}1$$$$f\left(x\right)=2x+3-\frac{x}{1-x^2}$$

Answered by mathmax by abdo last updated on 21/May/20

$$\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+3+\frac{\mathrm{x}}{{\mathrm{x}}^2-1}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to +\mathrm{\infty }}(2\mathrm{x}+3+\frac{1}{\mathrm{x}})$$$$=+\mathrm{\infty }$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{x}\right)=\frac{(2\mathrm{x}+3)(1-{\mathrm{x}}^2)-\mathrm{x}}{1-{\mathrm{x}}^2}=\frac{2\mathrm{x}-{2\mathrm{x}}^3+3-{3\mathrm{x}}^2-\mathrm{x}}{1-{\mathrm{x}}^2}$$$$=\frac{-{2\mathrm{x}}^3-{3\mathrm{x}}^2+\mathrm{x}+3}{1-{\mathrm{x}}^2}=\frac{{2\mathrm{x}}^3+{3\mathrm{x}}^2-\mathrm{x}-3}{(\mathrm{x}-1)(\mathrm{x}+1)}\Rightarrow$$$${\lim }_{\mathrm{x}\to 1^{+}}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{0^{+}}=+\mathrm{\infty }\mathrm{and}{\lim }_{\mathrm{x}\to 1^{-}}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{0^{-}}=-\mathrm{\infty }$$$${\lim }_{\mathrm{x}\to -1^{+}}\mathrm{f}\left(\mathrm{x}\right)=\frac{-1}{(-2)\times {\mathrm{o}}^{+}}=+\mathrm{\infty }$$$${\lim }_{\mathrm{x}\to -1^{-}}\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{\infty }$$

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