Evaluate the following limit.
\[
\lim _{x \rightarrow-\infty} \frac{-2 e^{-x}}{x^{2}+2 x-5}
\]

Recall that

Step 1 ：The limit is of the form 0/0 as x approaches -∞. This is an indeterminate form, so we can use L'Hopital's rule to evaluate the limit. L'Hopital's rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

Step 2 ：So, we need to find the derivative of the numerator and the denominator and then evaluate the limit again. If the new limit is still indeterminate, we can apply L'Hopital's rule again until we get a determinate form.

Step 3 ：The numerator is \(-2e^{-x}\) and its derivative is \(2e^{-x}\).

Step 4 ：The denominator is \(x^{2} + 2x - 5\) and its derivative is \(2x + 2\).

Step 5 ：Substituting these into the limit, we get \(\lim _{x \rightarrow-\infty} \frac{2e^{-x}}{2x + 2}\).

Step 6 ：The new limit is -∞, which is a determinate form. Therefore, we can conclude that the original limit is -∞.

Step 7 ：Final Answer: \(\boxed{-\infty}\)