Apply an exponential of a logarithm to the expression.
$\quad\lim\limits_{x\to \infty}(x^{2}+1)^{e^{-x}}=\lim\limits_{x\to \infty} \exp \left(\log ((x^{2}+1)^{e^{-x}})\right)$
$=\lim\limits_{x\to \infty} \exp \left(\dfrac{\log(x^{2}+1)}{e^{x}}\right)$
Since the exponential function is continuous, we may factor it out of the limit.
$\quad\lim\limits_{x\to \infty} \exp \left(\dfrac{\log(x^{2}+1)}{e^{x}}\right)$
$=\exp \left(\lim\limits_{x\to \infty} \dfrac{\log(x^{2}+1)}{e^{x}}\right)$
The numerator of $e^{-x} \log (x^{2}+1)$ grows asymptotically slower than its denominator as $x$ approaches $\infty$.
Since $\log (x^{2}+1)$ grows asymptotically slower than $e^{x}$ as $x$ approaches $\infty$, $\lim\limits_{x\to \infty}e^{-x} \log (x^{2}+1)=0 : e^{0}.$
Evaluate $e^{0}$.
$e^{0}=1$:
Answer: $1.$
Correct Answer: $C$