Question Number 10829 by Saham last updated on 26/Feb/17
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\int_{\:\mathrm{1}} ^{\:\mathrm{x}} \:\:\:\frac{\mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \:\left(\mathrm{dt}\right)}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{1}}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{1}\:\left(\mathrm{b}\right)\:\mathrm{0}\:\left(\mathrm{c}\right)\:\mathrm{e}/\mathrm{2}\:\left(\mathrm{d}\right)\:\mathrm{e} \\ $$
Answered by bahmanfeshki last updated on 27/Feb/17
$${f}\left({x}\right)=\int_{\mathrm{1}} ^{{x}} {e}^{{t}^{\mathrm{2}} } \:{dt}\Rightarrow{f}'\left({x}\right)={e}^{{x}^{\mathrm{2}} } \Rightarrow\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{1}} ^{{x}} {e}^{{t}^{\mathrm{2}} } \:{dt}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\int_{\mathrm{1}} ^{{x}} {e}^{{t}^{\mathrm{2}} } \:{dt}}{{x}−\mathrm{1}}\right)\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)={f}'\left(\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{2}}=\frac{{e}}{\mathrm{2}} \\ $$
Commented by FilupS last updated on 27/Feb/17
$$\mathrm{incorrect} \\ $$
Commented by bahmanfeshki last updated on 27/Feb/17
$${edited} \\ $$
Commented by Saham last updated on 27/Feb/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by FilupS last updated on 27/Feb/17
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\int_{\mathrm{1}} ^{\:{x}} \frac{{e}^{{t}^{\mathrm{2}} } }{{x}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{1}} ^{\:{x}} {e}^{{t}^{\mathrm{2}} } {dt} \\ $$$$\int{e}^{{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\mathrm{erf}\left({x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\left(\mathrm{erf}\left({x}\right)−\mathrm{erf}\left(\mathrm{1}\right)\right) \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{erf}\left({x}\right)−\mathrm{erf}\left(\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\mathrm{L}'\mathrm{Hopital}'\mathrm{s}\:\mathrm{Law} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left(\mathrm{erf}\left({x}\right)−\mathrm{erf}\left(\mathrm{1}\right)\right)}{\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\mathrm{erf}\left(\mathrm{1}\right)=\mathrm{constant} \\ $$$$\Rightarrow\mathrm{erf}\left({x}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\frac{\mathrm{2}}{\:\sqrt{\pi}}{e}^{{x}^{\mathrm{2}} } −\mathrm{0}}{\mathrm{2}{x}−\mathrm{0}}\right) \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{e}^{{x}^{\mathrm{2}} } }{\mathrm{2}{x}}\right) \\ $$$$\frac{{e}^{\mathrm{1}} }{\mathrm{2}} \\ $$$$\: \\ $$$$\therefore\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\int_{\mathrm{1}} ^{\:{x}} \frac{{e}^{{t}^{\mathrm{2}} } }{{x}^{\mathrm{2}} −\mathrm{1}}{dt}\:=\:\frac{{e}}{\mathrm{2}} \\ $$
Commented by Saham last updated on 27/Feb/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by bahmanfeshki last updated on 27/Feb/17
$${my}\:{answer}\:{is}\:{edited}… \\ $$