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Question Number 96931 by bobhans last updated on 05/Jun/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{5\mathrm{x}}^4-8}{{7\mathrm{x}}^3+2}\times \tan \left(\frac{3}{\mathrm{x}}\right)=?$$

Commented by PRITHWISH SEN 2 last updated on 05/Jun/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{5-\frac{8}{{\mathrm{x}}^4}}{7+\frac{2}{{\mathrm{x}}^3}}\left\{\underset{\frac{3}{\mathrm{x}}\to 0}{\lim }\frac{\tan \left(\frac{3}{\mathrm{x}}\right)}{\frac{3}{\mathrm{x}}}\times 3\right\}=\frac{5}{7}\times 1\times 3=\frac{15}{7}$$

Commented by bemath last updated on 06/Jun/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{15\mathrm{x}}^4-24}{{7\mathrm{x}}^4+2\mathrm{x}}\times \frac{\tan \left(\frac{3}{\mathrm{x}}\right)}{\left(\frac{3}{\mathrm{x}}\right)}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{15\mathrm{x}}^4-24}{{7\mathrm{x}}^4+2\mathrm{x}}\times 1=\frac{15}{7}$$$$$$

Commented by bobhans last updated on 06/Jun/20

$$\mathrm{yes}..\mathrm{thanks}\mathrm{all}$$

Answered by RAMANA last updated on 05/Jun/20

$$icouldnottypebuteasy$$

Answered by 1549442205 last updated on 05/Jun/20

$$\mathrm{putting}\mathrm{x}=\frac{1}{\mathrm{t}}\mathrm{we}\mathrm{have}\underset{\mathrm{t}\to 0}{\lim }\frac{\frac{5}{{\mathrm{t}}^4}-8}{\frac{7}{{\mathrm{t}}^3}+2}\times \tan \left(3\mathrm{t}\right)=$$$$\underset{\mathrm{t}\to 0}{\mathrm{Lim}}\frac{5-{8\mathrm{t}}^4}{{2\mathrm{t}}^4+7\mathrm{t}}\times \tan \left(3\mathrm{t}\right)=\underset{\mathrm{t}\to 0}{\mathrm{Lim}}\frac{5-{8\mathrm{t}}^4}{{2\mathrm{t}}^3+7}\times \frac{\sin \left(3\mathrm{t}\right)}{3\mathrm{t}}\times \frac{3}{\cos \left(3\mathrm{t}\right)}=$$$$\frac{5}{7}\times 1\times \frac{3}{1}=\frac{15}{7}$$

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