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Question Number 207233 by hardmath last updated on 10/May/24

$$1.\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\mathrm{n}-1}{\mathrm{n}+2}\right)}^{\mathbf{n}+3}=?$$$$$$$$2.\underset{\mathbf{x}\to \mathrm{\infty }}{\lim }{\left(\frac{5\mathrm{x}+6}{2\mathrm{x}-9}\right)}^{{\mathbf{x}}^2}=?$$$$$$$$3.\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\mathrm{n}+3}{\mathrm{n}+1}\right)}^{\mathbf{n}}=?$$

Commented by Frix last updated on 10/May/24

$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n+a}{n+b}\right)}^n={\mathrm{e}}^{a-b}$$

Answered by AliJumaa last updated on 10/May/24

$$1:$$$$letx-2=n\Rightarrow x\to \mathrm{\infty }$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }{(1-\frac{3}{x})}^{x+1}=e^{-3}$$$$2)$$$${\left(\underset{x\to \mathrm{\infty }}{\lim }\frac{5x+6}{2x-9}\right)}^{\underset{x\to \mathrm{\infty }}{\lim }}=\mathrm{\infty }$$$$3)$$$$let:x-1=n\Rightarrow x\to \mathrm{\infty }$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }{(1-\frac{2}{x})}^{x-1}=\mathrm{\infty }$$

Commented by Frix last updated on 10/May/24

$$3)\mathrm{is}\mathrm{wrong}.\mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}{\mathrm{e}}^2$$

Commented by hardmath last updated on 10/May/24

$$\mathrm{dear}\mathrm{professor},$$$$2.=\mathrm{\infty }?$$

Commented by Frix last updated on 10/May/24

$$\mathrm{Yes}.$$$$\frac{5x+6}{2x-9}=\frac{5}{2}+\frac{57}{2(2x-9)};{\left(\frac{5}{2}\right)}^{k\to \mathrm{\infty }}=\mathrm{\infty }$$$$$$$$\mathrm{In}\mathrm{the}\mathrm{other}\mathrm{cases}$$$$\frac{n+a}{n+b}=1+\frac{a-b}{n+b};1^{k\to \mathrm{\infty }}=1$$

Commented by AliJumaa last updated on 11/May/24

$$thankyouprofessorimateenmyageis17andiwanttobecomesomedayamathmatition$$

Commented by Frix last updated on 11/May/24

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{welcome}!$$$$\mathrm{Never}\mathrm{stop}\mathrm{learning}!$$

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