{\frac{{{{(1 + \tan x)}^{\frac{1}{x}}} - e}}{x};x \ne 0}\\
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;x \ne 0}
\end{array}} \right.$ be continuous at $x = 0$, then $k$ is-
- ✓$-\frac{e}{2}$
- B$-e$
- C$-\frac{e}{4}$
- D$\frac{e}{4}$
${\rm{k}} = e\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{{{\rm{e}}^{\frac{1}{{\rm{x}}}\ln (1 + \tan {\rm{x}}) - 1}} - 1}}{{\rm{x}}}} \right)$
${\rm{k}} = e\mathop {\lim }\limits_{{\rm{x}} \to 0} \left( {\frac{{{{\rm{e}}^{\frac{1}{{\rm{x}}}\ln (1 + \tan x) - 1}} - 1}}{{\frac{1}{{\rm{x}}}\ln (1 + \tan {\rm{x}}) - 1}}} \right)\left( {\frac{{\ln (1 + \tan {\rm{x}}) - {\rm{x}}}}{{{{\rm{x}}^2}}}} \right)$
$k=e \times\left(-\frac{1}{2}\right)=-\frac{e}{2}$
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${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then