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1-lim-n-n-1-n-2-n-3-2-lim-x-5x-6-2x-9-x-2-3-lim-n-n-3-n-1-n-

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Question Number 207233 by hardmath last updated on 10/May/24
1. lim_(n→∞) ( ((n − 1)/(n + 2)) )^(n + 3) = ? 2. lim_(x→∞) ( ((5x + 6)/(2x − 9)) )^x^2 = ? 3. lim_(n→∞) ( ((n + 3)/(n + 1)) )^n = ?
$$\mathrm{1}.\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{n}\:−\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}\:\right)^{\boldsymbol{\mathrm{n}}\:+\:\mathrm{3}} \:=\:\:? \\ $$$$ \\ $$$$\mathrm{2}.\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{5x}\:+\:\mathrm{6}}{\mathrm{2x}\:−\:\mathrm{9}}\:\right)^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:=\:\:? \\ $$$$ \\ $$$$\mathrm{3}.\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{n}\:+\:\mathrm{3}}{\mathrm{n}\:+\:\mathrm{1}}\:\right)^{\boldsymbol{\mathrm{n}}} \:=\:\:? \\ $$
Commented by Frix last updated on 10/May/24
lim_(n→∞) (((n+a)/(n+b)))^n =e^(a−b)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}+{a}}{{n}+{b}}\right)^{{n}} \:=\mathrm{e}^{{a}−{b}} \\ $$
Answered by AliJumaa last updated on 10/May/24
1: let x−2=n ⇒x→∞ ⇒ lim_(x→∞) (1−(3/x))^(x+1) =e^(−3) 2) (lim_(x→∞) ((5x+6)/(2x−9)))^(lim_(x→∞) ) =∞ 3) let: x−1=n ⇒x→∞ ⇒ lim_(x→∞) (1−(2/x))^(x−1) =∞
$$\mathrm{1}: \\ $$$${let}\:{x}−\mathrm{2}={n}\:\Rightarrow{x}\rightarrow\infty \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{3}}{{x}}\right)^{{x}+\mathrm{1}} ={e}^{−\mathrm{3}} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{5}{x}+\mathrm{6}}{\mathrm{2}{x}−\mathrm{9}}\right)^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}} =\infty \\ $$$$\left.\mathrm{3}\right) \\ $$$${let}:\:{x}−\mathrm{1}={n}\:\Rightarrow{x}\rightarrow\infty \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)^{{x}−\mathrm{1}} =\infty \\ $$
Commented by Frix last updated on 10/May/24
3) is wrong. The correct answer is e^2
$$\left.\mathrm{3}\right)\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{The}\:\mathrm{correct}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{e}^{\mathrm{2}} \\ $$
Commented by hardmath last updated on 10/May/24
dear professor, 2. = ∞ ?
$$\mathrm{dear}\:\mathrm{professor}, \\ $$$$\mathrm{2}.\:\:=\:\infty\:? \\ $$
Commented by Frix last updated on 10/May/24
Yes. ((5x+6)/(2x−9))=(5/2)+((57)/(2(2x−9))); ((5/2))^(k→∞) =∞ In the other cases ((n+a)/(n+b))=1+((a−b)/(n+b)); 1^(k→∞) =1
$$\mathrm{Yes}. \\ $$$$\frac{\mathrm{5}{x}+\mathrm{6}}{\mathrm{2}{x}−\mathrm{9}}=\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{57}}{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{9}\right)};\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{k}\rightarrow\infty} =\infty \\ $$$$ \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{other}\:\mathrm{cases} \\ $$$$\frac{{n}+{a}}{{n}+{b}}=\mathrm{1}+\frac{{a}−{b}}{{n}+{b}};\:\mathrm{1}^{{k}\rightarrow\infty} =\mathrm{1} \\ $$
Commented by AliJumaa last updated on 11/May/24
thank you professor im a teen my age is 17 and i want to become some day a mathmatition
$${thank}\:{you}\:{professor}\:{im}\:{a}\:{teen}\:{my}\:{age}\:{is}\:\mathrm{17}\:{and}\:{i}\:{want}\:{to}\:{become}\:{some}\:{day}\:{a}\:{mathmatition} \\ $$
Commented by Frix last updated on 11/May/24
You′re welcome! Never stop learning!
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}! \\ $$$$\mathrm{Never}\:\mathrm{stop}\:\mathrm{learning}! \\ $$

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