So I'm trying to do my homework on L'Hopital's rule. There's this one problem I have to do where I have to find the limit as x approaches negative infinity of (x^4)(e^x). How do I do this, since it's not in the form f(x)/g(x)?
Thanks
x^4*e^x = x^4/e^-x
use the Rule 4 times to get
24/e^-x = 24e^x --> 0
Oh I forgot I could use the rule more than once. Thanks!
To evaluate the given limit as x approaches negative infinity of (x^4)(e^x), you can indeed use L'Hopital's rule, even though it is not in the standard f(x)/g(x) form.
Here's how to apply L'Hopital's rule in this case:
1. Begin by noting that the function x^4 is a polynomial, so its derivative is 4x^3.
Similarly, e^x is an exponential function, and its derivative is e^x.
2. Rewrite the given limit in the form f(x)/g(x), where f(x) = x^4 and g(x) = e^(-x).
3. Now take the derivative of f(x) and g(x) separately. The derivative of f(x) = x^4 is f'(x) = 4x^3. The derivative of g(x) = e^(-x) is g'(x) = -e^(-x).
4. Apply L'Hopital's rule by calculating the limit of the ratio of the derivatives f'(x)/g'(x) as x approaches negative infinity.
lim (x→-∞) [f'(x) / g'(x)] = lim (x→-∞) [(4x^3) / (-e^(-x))]
5. Continuing with the limit, as x approaches negative infinity, the term 4x^3 goes to negative infinity since the leading term is x^3 and the coefficient is positive.
6. Similarly, the term e^(-x) goes to positive infinity as x approaches negative infinity.
7. In this case, the limit becomes:
lim (x→-∞) [4x^3 / -e^(-x)] = -∞ / ∞
8. The result of -∞ / ∞ is an indeterminate form. To resolve this issue, we can use L'Hopital's rule again.
9. Take the derivative of both f'(x) and g'(x). The derivative of f'(x) = 12x^2, and the derivative of g'(x) = e^(-x).
10. Now, evaluate the limit of the ratio of the second derivatives, f''(x)/g''(x), as x approaches negative infinity:
lim (x→-∞) [(12x^2) / (e^(-x))] = lim (x→-∞) [12x^2 / e^(-x)]
11. As x approaches negative infinity, the term 12x^2 still goes to negative infinity, and the term e^(-x) continues to go to positive infinity.
12. At this point, we again have an indeterminate form, -∞ / ∞. To resolve this, we repeat the process of taking the derivatives until we can evaluate the limit.
13. Take the derivative of both f''(x) and g''(x). The derivative of f''(x) = 24x, and the derivative of g''(x) = e^(-x).
14. Evaluate the limit of the ratio of the third derivatives, f'''(x)/g'''(x), as x approaches negative infinity:
lim (x→-∞) [(24x) / (e^(-x))] = lim (x→-∞) [24x / e^(-x)]
15. Again, as x approaches negative infinity, the term 24x goes to negative infinity, and the term e^(-x) goes to positive infinity.
16. We still have an indeterminate form, -∞ / ∞. To proceed, we can continue taking derivatives until we can evaluate the limit.
17. Take the derivative of both f'''(x) and g'''(x). The derivative of f'''(x) = 24, and the derivative of g'''(x) = e^(-x).
18. Evaluate the limit of the ratio of the fourth derivatives, f''''(x)/g''''(x), as x approaches negative infinity:
lim (x→-∞) [(24) / (e^(-x))] = 24 / e^(-∞)
19. At this stage, as x approaches negative infinity, e^(-x) approaches positive infinity. Therefore, e^(-∞) = 1 / e^∞ = 1 / 0 = ∞.
20. Finally, the result of the limit is 24 / ∞ = 0, since any finite number divided by infinity is equal to zero.
Therefore, the limit as x approaches negative infinity of (x^4)(e^x) is equal to zero.
Remember, L'Hopital's rule is most useful when we have indeterminate forms such as 0/0 or ∞/∞. However, if you encounter other forms like ∞ - ∞, 0 * ∞, or 1^∞, additional techniques may be necessary.