Question Number 75694 by malwaan last updated on 15/Dec/19
$$\boldsymbol{{prove}}\:\boldsymbol{{that}}\: \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\:\boldsymbol{{x}}^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} \:=\mathrm{1} \\ $$
Answered by vishalbhardwaj last updated on 15/Dec/19
$$\left({as}\:{x}\rightarrow\infty\:\mathrm{then}\:\frac{\mathrm{1}}{{x}}\:\rightarrow\:\mathrm{0}\right) \\ $$$$\Rightarrow\:{li}\underset{{x}\rightarrow\:\infty} {{m}x}^{{li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{1}}{{x}}} =\:{li}\underset{{x}\rightarrow\infty} {{m}}\left({x}\right)^{\mathrm{0}} \:=\:\mathrm{1} \\ $$
Commented by malwaan last updated on 15/Dec/19
$${but}\:\underset{{x}\rightarrow\infty} {{lim}}\:{x}^{\mathrm{0}} \:=\:\infty^{\mathrm{0}} \:{undefind} \\ $$
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
$$\mathrm{you}\:\mathrm{cannot}\:\mathrm{split}\:\mathrm{it}\:\mathrm{like}\:\mathrm{this} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\frac{\mathrm{1}}{{x}}} \neq\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\right)^{\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\mathrm{because}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\:=\infty\notin\mathbb{R} \\ $$$$\mathrm{and}\:{r}^{\mathrm{0}} =\mathrm{1}\:\mathrm{only}\:\mathrm{for}\:{r}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$ \\ $$
Answered by MJS last updated on 15/Dec/19
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\frac{\mathrm{1}}{{x}}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}e}^{\frac{\mathrm{ln}\:{x}}{{x}}} =\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:{x}}{{x}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:{x}}{{x}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left[\mathrm{ln}\:{x}\right]}{\frac{{d}}{{dx}}\left[{x}\right]}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:{x}}{{x}}} =\mathrm{1}\:\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}e}^{\frac{\mathrm{ln}\:{x}}{{x}}} =\mathrm{1}\:\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\frac{\mathrm{1}}{{x}}} =\mathrm{1} \\ $$
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
$${thank}\:{you}\:{so}\:{much} \\ $$
Answered by $@ty@m123 last updated on 15/Dec/19
$${Solution}: \\ $$$${Let}\:\:{y}=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:{x}^{\frac{\mathrm{1}}{{x}}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{ln}\:{y}=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{1}}{{x}}\mathrm{ln}\:{x} \\ $$$$\:\Rightarrow\:\:\:\mathrm{ln}\:{y}=\:\mathrm{0} \\ $$$$\:\Rightarrow\:{y}=\mathrm{1}\:\:\:\:\:\:\: \\ $$