The value of $$\mathop {\lim }\limits_{{\text{x}} \to \infty } {\left( {1 + {{\text{x}}^2}} \right)^{{{\text{e}}^{ – {\text{x}}}}}}$$ is A. 0 B. $$\frac{1}{2}$$ C. 1 D. $$\infty $$

The correct answer is $\boxed{\infty}$.

Consider the following graph of the function $y = \left( {1 + {{\text{x}}^2}} \right)^{{{\text{e}}^{ – {\text{x}}}}}$:

[asy]
unitsize(1 cm);

draw((0,0)–(10,0));
draw((0,0)–(0,1));

real ticklen=1.2;
real tickspace=2;
real axisarrowsize=0.14inch;
real tickdown=-0.15inch;
real tickdownlength=-0.12inch;
real wholetickdown=tickdown;
real wholetickdownlength=tickdownlength;
real tickdownbase=0.1inch;
real wholetickdownbase=tickdownbase;
real t=0;
real[] TicksArrx,TicksArry;

for (int i=-5; i<=5; ++i) {
TicksArrx.push(t);
TicksArry.push(exp(-t)exp(1+ii));
t+=tickspace;
}

xaxis(BottomTop, Ticks(“%”, TicksArrx, pTick=gray(0.22),extend=true),p=invisible);
yaxis(LeftRight, Ticks(“%”, TicksArry, pTick=gray(0.22),extend=true),p=invisible);

label(“$x$”,(10,0),SE);
label(“$y$”,(0,1),NW);

dot(“$(0,1)$”, (0,1), S);
[/asy]

As $x$ approaches $\infty$, the value of $y$ approaches $\infty$. This is because the exponential function $e^x$ grows very quickly, and as $x$ gets larger, the value of $e^{-x}$ gets smaller and smaller. Therefore, the product of $e^{-x}$ and any other number, such as $1 + x^2$, approaches 0 as $x$ approaches $\infty$.

The other options are incorrect because they do not represent the correct behavior of the function as $x$ approaches $\infty$. Option A, 0, is incorrect because the value of $y$ does not approach 0 as $x$ approaches $\infty$. Option B, $\frac{1}{2}$, is incorrect because the value of $y$ does not approach $\frac{1}{2}$ as $x$ approaches $\infty$. Option C, 1, is incorrect because the value of $y$ does not approach 1 as $x$ approaches $\infty$.