Question Number 211717 by liuxinnan last updated on 18/Sep/24
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{li}{m}}\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}}{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}=? \\ $$
Answered by BHOOPENDRA last updated on 18/Sep/24
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{2}{x}}{\mathrm{3}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }\right)}{\left(\mathrm{2}{xe}^{{x}^{\mathrm{2}} } \right)}\:\:{Apply}\:{L}'{H}\hat {{o}pital}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} {e}^{{x}^{\mathrm{2}} } } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{2}/\mathrm{3}} {e}^{\mathrm{0}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$