\( \lim _ { x \rightarrow - \infty } \frac { x ^ { 2 } + 1 } { e ^ { x } } \)

To evaluate the limit \( \lim _ { x \rightarrow - \infty } \frac { x ^ { 2 } + 1 } { e ^ { x } } \), we first notice that as \( x \) approaches \(-\infty\), \( e^x \) approaches 0, while \( x^2 + 1 \) approaches infinity. This suggests that the growth of the numerator is outpaced by the rapid decay of the denominator. We can analyze this formally: 1. Focus on the term in the numerator, \( x^2 + 1 \), which behaves like \( x^2 \) for large values of \( |x| \). 2. Rewrite the limit as \( \lim _ { x \rightarrow -\infty } \frac { x^2 } { e^x } \). Now, applying L'Hôpital's Rule since both the numerator and the denominator tend to infinity: - Differentiate the numerator: \( \frac{d}{dx}(x^2) = 2x \). - Differentiate the denominator: \( \frac{d}{dx}(e^x) = e^x \). Now we have: \[ \lim _ { x \rightarrow -\infty } \frac { 2x } { e^x } \] This limit, in itself, again yields an indeterminate form \(-\infty / 0\), so we apply L'Hôpital's Rule once more: - Differentiate the numerator again: \( \frac{d}{dx}(2x) = 2 \). - Differentiate the denominator again: \( \frac{d}{dx}(e^x) = e^x \). Now we compute: \[ \lim _ { x \rightarrow -\infty } \frac { 2 } { e^x } = 0 \] Thus, returning to our original limit, we conclude: \[ \lim _ { x \rightarrow -\infty } \frac { x ^ { 2 } + 1 } { e ^ { x } } = 0. \]