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Question Number 22365 by A1B1C1D1 last updated on 16/Oct/17

Answered by sma3l2996 last updated on 16/Oct/17

=lim_(x→∞) ((1−((3/5))^x +(1/5^x ))/(1+((3/5))^x +(1/(5^x x))))=((1−0+0)/(1+0+0))=1

$$=\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} {x}}}=\frac{\mathrm{1}−\mathrm{0}+\mathrm{0}}{\mathrm{1}+\mathrm{0}+\mathrm{0}}=\mathrm{1}\:\: \\ $$

Commented by A1B1C1D1 last updated on 16/Oct/17

Sorry, but which method did you use?

$$\mathrm{Sorry},\:\mathrm{but}\:\mathrm{which}\:\mathrm{method}\:\mathrm{did}\:\mathrm{you}\:\mathrm{use}? \\ $$

Commented by sma3l2996 last updated on 16/Oct/17

lim_(x→∞) ((5^x −3^x +1)/(5^x +3^x +(1/x)))=lim_(x→∞) ((5^x (1−(3^x /5^x )+(1/5^x )))/(5^x (1+(3^x /5^x )+(1/(5^x ×x))))) =lim_(x→∞) ((1−((3/5))^x +(1/5^x ))/(1+((3/5))^x +(1/(5^x ×x))))=((1−((3/5))^∞ +5^(−∞) )/(1+((3/5))^∞ +(1/(∞×e^∞ ))))=((1−0+0)/(1+0+0))=1 lim_(x→∞) ((3/5))^x =0 because (3/5)<1 and lim_(x→∞) 5^x x=lim_(x→∞) xe^(xln5) =∞

$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{5}^{{x}} −\mathrm{3}^{{x}} +\mathrm{1}}{\mathrm{5}^{{x}} +\mathrm{3}^{{x}} +\frac{\mathrm{1}}{{x}}}=\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{5}^{{x}} \left(\mathrm{1}−\frac{\mathrm{3}^{{x}} }{\mathrm{5}^{{x}} }+\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\right)}{\mathrm{5}^{{x}} \left(\mathrm{1}+\frac{\mathrm{3}^{{x}} }{\mathrm{5}^{{x}} }+\frac{\mathrm{1}}{\mathrm{5}^{{x}} ×{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} ×{x}}}=\frac{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\infty} +\mathrm{5}^{−\infty} }{\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\infty} +\frac{\mathrm{1}}{\infty×{e}^{\infty} }}=\frac{\mathrm{1}−\mathrm{0}+\mathrm{0}}{\mathrm{1}+\mathrm{0}+\mathrm{0}}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} =\mathrm{0}\:\:\:{because}\:\:\frac{\mathrm{3}}{\mathrm{5}}<\mathrm{1} \\ $$$${and}\:\:\underset{{x}\rightarrow\infty} {{lim}}\mathrm{5}^{{x}} {x}=\underset{{x}\rightarrow\infty} {{lim}xe}^{{xln}\mathrm{5}} =\infty \\ $$

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