How does lim as t-->-∞ (t/e^-t) becomes lim as t-->-∞ (1/(-1)e^-t). Where does does -1 comes from?
Recall the chain rule: if u is a function of t, then
the derivative of e^u = e^u du/dt
so, d/dt (e^-t) = -1 e^-t
Thank you
To understand how the expression changes from lim as t approaches negative infinity of (t/e^-t) to lim as t approaches negative infinity of (1/(-1)e^-t), let's break it down step by step:
1. Start with the original expression: lim as t approaches negative infinity of (t/e^-t).
2. Since we have e^-t in the denominator, which approaches 0 as t approaches negative infinity, we can apply L'Hospital's rule. This rule states that if the limit of a fraction of two functions exists in an indeterminate form (0/0 or infinity/infinity), then taking the derivative of both the numerator and denominator can help simplify the expression.
3. Apply L'Hospital's rule by taking the derivatives of both the numerator and denominator with respect to t. The derivative of t is 1, and the derivative of e^-t is -e^-t.
4. After taking the derivatives, we get lim as t approaches negative infinity of (1/(-e^-t)).
5. To simplify further, we multiply the numerator and denominator by e^t, which does not change the value of the expression: lim as t approaches negative infinity of ((e^t)/(e^t * -e^-t)).
6. Simplify the denominator: e^t * -e^-t = -e^(-t + t) = -e^0 = -1.
7. Thus, the expression becomes lim as t approaches negative infinity of ((e^t)/(-1)).
8. Finally, recognize that e^t approaches infinity as t approaches negative infinity. So, the final expression is lim as t approaches negative infinity of (∞/(-1)).
9. Since we have a fraction with infinity in the numerator and -1 in the denominator, the limit evaluates to -∞.
Therefore, the expression lim as t approaches negative infinity of (t/e^-t) becomes lim as t approaches negative infinity of (1/(-1)e^-t), and the -1 comes from simplifying the denominator using L'Hospital's rule and algebraic manipulation.