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Question Number 107993 by mohammad17 last updated on 13/Aug/20

Answered by hgrocks last updated on 13/Aug/20

$$\mathrm{\infty }$$

Commented by mohammad17 last updated on 13/Aug/20

$$siriwantstebbysteb$$

Answered by hgrocks last updated on 13/Aug/20

$$\mathrm{cosx}-\mathrm{x}!=-\mathrm{x}!\mathrm{as}\mathrm{x}\to \mathrm{\infty }$$$$3^{\mathrm{x}}-4^{\mathrm{x}}=4^{\mathrm{x}}(1-{\left(\frac{3}{4}\right)}^{\mathrm{x}})$$$$=-4^{\mathrm{x}}\{\mathrm{as}\mathrm{x}\to \mathrm{\infty }{\left(\frac{3}{4}\right)}^{\mathrm{x}}=0\}$$$$\mathrm{So}\mathrm{L}={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{-\mathrm{x}!}{-4^{\mathrm{x}}}$$$$=\frac{1}{4}.\frac{2}{4}.\frac{3}{4}.\frac{4}{4}............\frac{\mathrm{x}}{4}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}{\left(\frac{\mathrm{x}}{4\mathrm{e}}\right)}^{\mathrm{x}}\sqrt{2\pi \mathrm{x}}$$$$=\mathrm{\infty }$$

Commented by mohammad17 last updated on 13/Aug/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 13/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cosx}-\mathrm{x}!}{3^{\mathrm{x}}-4^{\mathrm{x}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cosx}-\mathrm{x}!}{4^{\mathrm{x}}({\left(\frac{3}{4}\right)}^{\mathrm{x}}-1)}=\frac{\mathrm{x}!(\frac{\mathrm{cosx}}{\mathrm{x}!}-1)}{4^{\mathrm{x}}({\left(\frac{3}{4}\right)}^{\mathrm{x}}-1)}$$$$\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{-\mathrm{x}!}{-4^{\mathrm{x}}}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}}{4^{\mathrm{x}}}$$$$={\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\left(\frac{\mathrm{x}}{4}\right)}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{e}}^{\mathrm{xln}\left(\frac{\mathrm{x}}{4}\right)-\mathrm{x}}\sqrt{2\pi \mathrm{x}}=+\mathrm{\infty }$$

Commented by mohammad17 last updated on 13/Aug/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 13/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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