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Question Number 75694 by malwaan last updated on 15/Dec/19

$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$\underset{\mathit{x}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}{\mathit{x}}^{\frac{1}{\mathit{x}}}=1$$

Answered by vishalbhardwaj last updated on 15/Dec/19

$$(asx\to \mathrm{\infty }\mathrm{then}\frac{1}{x}\to 0)$$$$\Rightarrow li{\underset{x\to \mathrm{\infty }}{mx}}^{li\underset{x\to \mathrm{\infty }}{m}\frac{1}{x}}=li\underset{x\to \mathrm{\infty }}{m}{\left(x\right)}^0=1$$

Commented by malwaan last updated on 15/Dec/19

$$but\underset{x\to \mathrm{\infty }}{lim}{x}^0={\mathrm{\infty }}^{0}undefind$$

Commented by vishalbhardwaj last updated on 15/Dec/19

$$\mathrm{whatever}\mathrm{the}\mathrm{value}\mathrm{if}x\mathrm{is}\mathrm{there},\mathrm{but}$$$$\mathrm{the}\mathrm{whole}\mathrm{power}\mathrm{is}\mathrm{zero},\mathrm{that}\mathrm{why}$$$$\mathrm{its}\mathrm{value}\mathrm{will}\mathrm{be}\mathrm{zero}\mathrm{in}\mathrm{the}\mathrm{sense}$$

Commented by MJS last updated on 15/Dec/19

$$\mathrm{you}\mathrm{cannot}\mathrm{split}\mathrm{it}\mathrm{like}\mathrm{this}$$$$\underset{x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{x}}\ne {\left(\underset{x\to \mathrm{\infty }}{\lim }x\right)}^{\left(\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x}\right)}$$$$\mathrm{because}\underset{x\to \mathrm{\infty }}{\lim }x=\mathrm{\infty }\notin \mathbb{R}$$$$\mathrm{and}{r}^0=1\mathrm{only}\mathrm{for}r\in \mathbb{R}\mathrm{\setminus }\left\{0\right\}$$$$$$

Answered by MJS last updated on 15/Dec/19

$$\underset{x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{x}}={\underset{x\to \mathrm{\infty }}{\lim e}}^{\frac{\ln x}{x}}={\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln x}{x}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln x}{x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{d}{dx}\left[\ln x\right]}{\frac{d}{dx}\left[x\right]}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x}=0$$$$\Rightarrow {\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln x}{x}}=1\Rightarrow {\underset{x\to \mathrm{\infty }}{\lim e}}^{\frac{\ln x}{x}}=1\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{x}}=1$$

Commented by vishalbhardwaj last updated on 15/Dec/19

$$\mathrm{sir},\mathrm{Is}\mathrm{my}\mathrm{assumption}\mathrm{and}\mathrm{explantion}\mathrm{wrong}??$$

Commented by malwaan last updated on 16/Dec/19

$$thankyousomuch$$

Answered by $@ty@m123 last updated on 15/Dec/19

$$Solution:$$$$Lety=\underset{x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{x}}$$$$$$$$\ln y=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x}\ln x$$$$\Rightarrow \ln y=0$$$$\Rightarrow y=1$$

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